RD Sharma XI 2018 Solutions for Class 12 Science Math Chapter 16 Permutations are provided here with simple step-by-step explanations. These solutions for Permutations are extremely popular among class 12 Science students for Math Permutations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2018 Book of class 12 Science Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2018 Solutions. All RD Sharma XI 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 16.14:

Question 1:

In a class there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function. In how many ways can the teacher make this selection?

Answer:

No. of boys in the class = 27
No. of girls in the class = 14
Ways to select a boy = 27
Similarly, ways to select a girl = 14
∴ Number of ways to select 1 boy and 1 girl = 27× 14 = 378
  

Page No 16.14:

Question 2:

A person wants to buy one fountain pen, one ball pen and one pencil from a stationery shop. If there are 10 fountain pen varieties, 12 ball pen varieties and 5 pencil varieties, in how many ways can he select these articles?

Answer:

Number of fountain pen varieties = 10
Number of  ball pen varieties = 12
Number of pencil varieties = 5
Ways to select a fountain pen = 10
Ways to select a ball pen = 12
Ways to select a pencil = 5

Ways to select a fountain pen, a ball pen and a pencil = 10 ×12 × 5 = 600                (Using the fundamental principle of multiplication)

Page No 16.14:

Question 3:

From Goa to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes; air, rail and road. From Goa to Delhi via Bombay, how many kinds of routes are there?

Answer:

Number of routes from Goa to Bombay = 2
Number of routes from Bombay to Delhi = 3
Using fundamental principle of multiplication:
Number of routes from Goa to Delhi via Bombay = 2 × 3 = 6

Page No 16.14:

Question 4:

A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of calendars should it prepare to serve for all the possibilities in future years?

Answer:

The first day of the year can be any one of the days of the week, i.e the first day can be selected in 7 ways.
But, the year could also be a leap year.
So, the mint should prepare 7 calendars for the non-leap year and 7 calendars for the leap year.
So, total number of calendars that should be made = 7 + 7 = 14

Page No 16.14:

Question 5:

There are four parcels and five post-offices. In how many different ways can the parcels be sent by registered post?

Answer:

Number of ways of sending 1 parcel via registered post = 5
Number of ways of sending 4 parcels via registered post through 5 post offices = 5×5×5×5 = 625



Page No 16.15:

Question 6:

A coin is tossed five times and outcomes are recorded. How many possible outcomes are there?

Answer:

Number of outcomes when the coin is tossed for the first time = 2
Number of outcomes when the coin is tossed for the second time = 2
Thus, there would be 2 outcomes, each time the coin is tossed.
Total number of possible outcomes on tossing the coin five times = 2×2×2×2×2 = 32

Page No 16.15:

Question 7:

In how many ways can an examinee answer a set of ten true/false type questions?

Answer:

Number of ways of answering the first question = 2 (either true or false)
Similarly, each question can be answered in 2 ways.
∴ Total number of ways of answering all the 10 questions = 2×2×2×2×2×2×2×2×2×2 = 210 = 1024

Page No 16.15:

Question 8:

A letter lock consists of three rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock?

Answer:

Number of ways of marking each of the ring = 10 different letters
∴ Total number of ways of marking any letter on these three rings = 10×10×10 = 1000
Out of these 1000 combinations of the lock, 1 combination will be successful.
∴ Total number of unsuccessful attempts = 1000 -1 = 999

Page No 16.15:

Question 9:

There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each?

Answer:

Number of ways of answering the first three questions = 4 each
Number of ways of answering the remaining three questions = 2 each
∴ Total number of ways of answering all the questions = 4×4×4×2×2×2 = 512 

Page No 16.15:

Question 10:

There are 5 books on Mathematics and 6 books on Physics in a book shop. In how many ways can a students buy : (i) a Mathematics book and a Physics book (ii) either a Mathematics book or a Physics book?

Answer:

  Number of  books on mathematics = 5
  Number of books on physics = 6
   Number of ways of buying a mathematics book = 5
   Similarly, number of ways of buying a physics book = 6
  (i) By using fundamental principle of multiplication:
  Number of ways of buying a mathematics and a physics book = 6×5 = 30
  (ii) By using the fundamental principle of addition:
  Number of ways of buying either a mathematics or a physics book = 6 + 5 = 11

Page No 16.15:

Question 11:

Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags, one below the other?

Answer:

Number of flags = 7
∴ Number of ways of selecting one flag = 7
Number of ways of selecting the other flag = 6 (as only 6 colours are available for use)
A signal requires use of two flags
∴ Total number of signal that can be generated = 7×6 = 42

Page No 16.15:

Question 12:

A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?

Answer:

A boy can be selected from the first team in 6 ways and from the second team in 5 ways.
∴ Number of ways of arranging a match between the boys of the two teams = 6×5 = 30
Similarly, A girl can be selected from the first team in 4 ways and from the second team in 3 ways.
∴ Number of ways of arranging a match between the girls of the two teams = 4×3= 12
∴ Total number of matches = 30 + 12 = 42

Page No 16.15:

Question 13:

Twelve students complete in a race. In how many ways first three prizes be given?

Answer:

Number of competitors in the race = 12
Number of  competitors who can come first in the race = 12
Number of  competitors who can come second in the race = 11    (as one competitor has already come first in the race)
Number of  competitors who can come third in the race = 10
∴ Total number of ways of awarding the first three prizes = 12×11×10 = 1320

Page No 16.15:

Question 14:

How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?

Answer:

Number of ways of selecting the first term from the set {1, 2, 3} = 3
Corresponding to each of the selected first terms, the number of ways of selecting the common difference  from the set {1, 2, 3, 4, 5} = 5
∴ Total number of AP's that can be formed = 3×5 = 15

Page No 16.15:

Question 15:

From among the 36 teachers in a college, one principal, one vice-principal and the teacher-incharge are to be appointed. In how many ways can this be done?

Answer:

Total number of teachers in the college = 36
Number of ways of  selecting a principal = 36
Number of ways of selecting a vice-principal = 35 (as one of the teacher is already being selected for the post of principal)
Similarly, number of ways of selecting the teacher-incharge = 34
∴ Total number of ways of selecting all the three = 36×35×34 = 42840

Page No 16.15:

Question 16:

How many three-digit numbers are there with no digit repeated?

Answer:

The thousand's place cannot be zero.
∴ Number of ways of selecting the thousand's digit = 9
Number of ways of selecting the ten's digit = 9 ( as repetition of digits is not allowed and one digit has already been used in the thousand's place)
Similarly, number of ways of selecting the unit's digit = 8 (as two digits have been used for the thousand's and the ten's places)
∴ Total three digit number that can be formed = 9×9×8 = 648

Page No 16.15:

Question 17:

How many three-digit numbers are there?

Answer:

Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
Since the thousand's place cannot be zero, available digits to fill the thousand's place = 9
Number of ways of filling the ten's digit = 10
Similarly, number of ways of filling the unit's digit = 10
∴ Total number of three digit numbers = 9×10×10 = 900

Page No 16.15:

Question 18:

How many three-digit odd numbers are there?

Answer:

Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
Since the thousand's place cannot be zero, the number of ways of filling the thousand's place is 9.
Number of ways of filling the ten's place = 10
Number of ways of filling the unit's place = 5  {1, 3, 5, 7, 9}
Total 3-digit odd numbers = 9×10×5 = 450

Page No 16.15:

Question 19:

How many different five-digit number licence plates can be made if
(i) first digit cannot be zero and the repetition of digits is not allowed,
(ii) the first-digit cannot be zero, but the repetition of digits is allowed?

Answer:

(i) Since the first digit cannot be zero, the number of ways of filling the first digit = 9
  Number of ways of filling the second digit = 9     (Since repetition is not allowed)
  Number of ways of filling the third digit = 8
  Number of ways of filling the fourth digit = 7
  Number of ways of filling the fifth digit = 6
  Total number of licence plates that can be made = 9×9×8×7×6 = 27216

(ii) Since the first digit cannot be zero, the number of ways of filling the first digit = 9
      Number of ways of filling the second digit = 10    (Since repetition is allowed)
      Number of ways of filling the third digit = 10
      Number of ways of filling the fourth digit = 10
     Number of ways of filling the fifth digit = 10
     Total number of licence plates that can be made = 9×10×10×10×10 = 90000

Page No 16.15:

Question 20:

How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed?

Answer:

Since the  number has to be greater than 7000, the thousand's place can only be filled by three digits, i.e. 7, 8 and 9.
Now, the hundred's place can be filled with the remaining 4 digits as the repetition of the digits is not allowed.
The ten's place can be filled with the remaining 3 digits.
The unit's place can be filled with the remaining 2 digits.
Total numbers that can be formed = 3×4×3×2 = 72

Page No 16.15:

Question 21:

How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?

Answer:

Since the  number has to be greater than 8000, the thousand's place can be filled by only two digits, i.e. 8 and 9.
Now, the hundred's place can be filled with the remaining 4 digits as the repetition of the digits is not allowed.
The ten's place can be filled with the remaining 3 digits.
The unit's place can be filled with the remaining 2 digits.
Total numbers that can be formed = 2×4×3×2 = 48

Page No 16.15:

Question 22:

In how many ways can six persons be seated in a row?

Answer:

Number of seats available to the first person = 6
Number of seats available to the second person = 5
Number of seats available to the third person = 4
Number of seats available to the fourth person = 3
Number of seats available to the fifth person = 2
Number of seats available to the sixth person = 1
Total number of ways of making the seating arrangement = 6×5×4×3×2×1 = 720

Page No 16.15:

Question 23:

How many 9-digit numbers of different digits can be formed?

Answer:

Since the first digit cannot be zero, number of ways of filling the first digit = 9
Number of ways of filling the second digit = 9    (as repetition is not allowed or the digits are distinct)
Number of ways of filling the third digit = 8
Number of ways of filling the fourth digit = 7
Number of ways of filling the fifth digit = 6
Number of ways of filling the sixth digit = 5
Number of ways of filling the seventh digit = 4
Number of ways of filling the eighth digit = 3
Number of ways of filling the ninth digit = 2
Total such 9-digit numbers =  9×9×8×7×6×5×4×3×2 = 9 (9!)

Page No 16.15:

Question 24:

How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?

Answer:

Since the number is less than 1000, it could be a three-digit, two-digit or single-digit number.
Case I: Three-digit number:
Now, the hundred's place cannot be zero. Thus, it can be filled with three digits, i.e. 3, 5 and 7.
Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the hundred's place.
Thus, the unit's place can be filled by only 2 digits.
Number of ways of filling the ten's digit = 2  (as repetition is not allowed)
Total three-digit numbers that can be formed = 3×2×2 = 12

Case II: Two-digit number:
Now, the ten's place cannot be zero. Thus, it can be filled with three digits, i.e. 3, 5 and 7.
Also, the unit's place cannot be zero. This is because it is an odd number and one digit has already been used to fill the ten's place,
Thus, the unit's place can be filled by only 2 digits.
Total two-digit numbers that can be formed = 3×2 = 6

Case III: Single-digit number:
  It could be 3, 5 and 7.
Total single-digit numbers that can be formed = 3
Hence, required number = 12 + 6 + 3 = 21

Page No 16.15:

Question 25:

How many 3-digit numbers are there, with distinct digits, with each digit odd?

Answer:

The hundred's place can be filled by {1, 3, 5, 7, 9), i.e. 5 digits.
The ten's place can now be filled by 4 digits (as one digit is already used in the hundred's place and repetition is not allowed )
Similarly, the unit's place can be filled by 3 digits.
Total number of 3-digit numbers = 5×4×3 = 60

Page No 16.15:

Question 26:

How many different numbers of six digits each can be formed from the digits 4, 5, 6, 7, 8, 9 when repetition of digits is not allowed?

Answer:

Number of ways of filling the first digit = 6
Number of ways of filling the second digit = 5          (as repetition is not allowed)
Number of ways of filling the third digit = 4
Number of ways of filling the fourth digit =3
Number of ways of filling the fifth digit = 2
Number of ways of filling the sixth digit = 1
Total numbers = 6×5×4×3×2×1 = 720

Page No 16.15:

Question 27:

How many different numbers of six digits can be formed from the digits 3, 1, 7, 0, 9, 5 when repetition of digits is not allowed?

Answer:

The first digit cannot be zero. Thus, the first digit can be filled in 5 ways.
Number of ways for filling the second digit = 5    (as repetition of digits is not allowed)
Number of ways for filling the third digit = 4
Number of ways for filling the fourth digit = 3
Number of ways for filling the fifth digit = 2
Number of ways for filling the sixth digit = 1
Total numbers = 5×5×4×3×2×1 = 600

Page No 16.15:

Question 28:

How many four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?

Answer:

As the number has to be greater than 5000, the first digit can either be 5 or 9.
Hence, it can be filled only in two ways.
Number of ways for filling the second digit = 4
Number of ways for filling the third digit = 3           (as repetition is not allowed)
Number of ways for filling the fourth digit = 2
Total numbers = 2×4×3×2 = 48

Page No 16.15:

Question 29:

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?

Answer:

Number of ways of selecting the first letter = 6
Number of ways of selecting the second letter = 5        (as repetition of letters is not allowed)
Number of ways of selecting the digit in the third place = 10
Number of ways of selecting the digit in the fourth place = 9         (as repetition of digits is not allowed)
Number of ways of selecting the digit in the fifth place = 8
Number of ways of selecting the digit in the sixth place = 7
Possible serial numbers = 6×5×10×9×8×7 = 151200



Page No 16.16:

Question 30:

A number lock on a suitcase has 3 wheels each labelled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock.

Answer:

The digits in the sequence do not repeat.
 Number of ways of selecting the first digit = 10
 Number of ways of selecting the second digit = 9
 Number of ways of selecting the third digit = 8
Total number of possible sequences = 10×9×8 = 720
Of all the possible sequences, only one sequence is successful.
∴ Number of unsuccessful sequences = 720 - 1 = 719

Page No 16.16:

Question 31:

A customer forgets a four-digits code for an Automatic Teller Machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6 and 9. Find the largest possible number of trials necessary to obtain the correct code.

Answer:

Assuming that the code of an ATM  has all distinct digits.
Number of ways for selecting the first digit = 4
Number of ways for selecting the second digit = 3
Number of ways for selecting the third digit = 2
Number of ways for selecting the fourth digit = 1
Total number of possible codes for the ATM  = 4×3×2×1 = 24

Page No 16.16:

Question 32:

In how many ways can three jobs I, II and III be assigned to three persons A, B and C if one person is assigned only one job and all are capable of doing each job?

Answer:

Number of ways of assigning a job to person A = 3
Number of ways of assigning the remaining jobs to person B = 2               (since one job has already been assigned to person A)
                                                                                                                 
The number of ways of assigning the remaining job to person C = 1
Total number of ways of job assignment = 3×2×1 = 6

Page No 16.16:

Question 33:

How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4, if the digits can repeat?

Answer:

Case I: Four-digit number
Total number of ways in which the 4 digit number can be formed = 4×4×4×4 = 256

Now, the number of ways in which the 4-digit numbers greater than 4321 can be formed is as follows:
Suppose, the thousand's digit is 4 and hundred's digit is either 3 or 4.
∴ Number of ways = 2×4×4 = 32
But 4311, 4312, 4313, 4314, 4321 (i.e. 5 numbers) are less than or equal to 4321.
∴ Remaining number of ways = 256-32-5=229

Case II: Three-digit number
The hundred's digit can be filled in 4 ways.
Similarly, the ten's digit and the unit's digit can also be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of three-digit number = 4×4×4 = 64

Case III: Two-digit number
The ten's digit and the unit's digit can be filled in 4 ways each. This is because the repetition of  digits is allowed.
∴ Total number of two digit numbers = 4×4 = 16

Case IV: One-digit number
Single digit number can only be four.
∴ Required numbers = 229 + 64 + 16 +4 = 313

Page No 16.16:

Question 34:

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7 and 9 when no digit is repeated? How many of them are divisible by 10?

Answer:

The first digit of the number cannot be zero. Thus, it can be filled in 5 ways.
The number of ways of filling the second digit = 5        (as the repetition of digits is not allowed)
The number of ways of filling the third digit = 4
The number of ways of filling the fourth digit = 3
The number of ways of filling the fifth digit = 2
The number of ways of filling the sixth digit = 1
∴ Required numbers = 5×5×4×3×2×1 = 600

For the number to be divisible by 10, the sixth digit has to be zero.
Now, the first digit can be filled in 5 ways.
Number of ways of filling the second digit = 4
Number of ways of filling the third digit = 3
Number of ways of filling the fourth digit = 2
Number of ways of filling the fifth digit = 1
Number of ways of filling the sixth digit = 1
Total numbers divisible by 10 = 5×4×3×2×1×1 = 120

Page No 16.16:

Question 35:

If three six faced die each marked with numbers 1 to 6 on six faces, are thrown find the total number of possible outcomes.

Answer:

Number of possible outcomes on one dice = 6   {1,2,3,4,5,6}
Number of possible outcomes on both the other two dice = 6
∴ Total number of outcomes when three dice are thrown = 6×6×6 = 216

Page No 16.16:

Question 36:

A coin is tossed three times and the outcomes are recorded. How many possible outcomes are there? How many possible outcomes if the coin is tossed four times? Five times? n times?

Answer:

Total number of outcomes when a coin is tossed once = 2 (Heads, Tails)
Number of outcomes when the coin is tossed for the second time = 2
∴  Number of outcomes when the coin is tossed thrice = 2×2×2 = 8
Similarly, the number of outcomes when the coin is tossed four times = 2×2×2×2 = 16
Similarly, the number of outcomes when the coin is tossed five times  = 2×2×2×2×2 = 32
Similarly, the number of outcomes when the coin is tossed 'n' times  = 2×2×.....n times = 2n

Page No 16.16:

Question 37:

How many numbers of four digits can be formed with the digits 1, 2, 3, 4, 5 if the digits can be repeated in the same number?

Answer:

The thousand's place can be filled by any of the 5 digits.
∴ Number of ways of filling the thousand's place = 5
Since the digits can repeat in the number, the hundred's place, the ten's place and the unit's place can each be filled in 5 ways.
∴ Total numbers = 5×5×5×5 = 625

Page No 16.16:

Question 38:

How many three digit numbers can be formed by using the digits 0, 1, 3, 5, 7 while each digit may be repeated any number of times?

Answer:

Since the hundred's place cannot be zero, it can be filled by any of the 4 digits (1, 3, 5 and 7).
∴ Number of ways of filling the hundred's place = 4
Since the digits can be repeated in the number, the ten's place and the unit's place can each be filled in 5 ways.
∴ Total numbers = 4×5×5 = 100

Page No 16.16:

Question 39:

How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times?

Answer:

Since the number is less than 1000, it means that it is a three-digit number, a two-digit number or a single-digit number.
Three-digit numbers:
The hundred's place can be filled by 5 digits neglecting zero as it can't be zero.
The ten's place and the unit's place can be filled by 6  digits.
So, total number of three digit numbers = 5×6×6 = 180

Two-digit numbers:
The ten's place can be filled by 5 digits, except zero.

The unit's digit can be filled by 6 digits.
Total two digit numbers = 5×6 = 30

Single digit numbers are 1, 2, 3, 4, 5 as 0 is not a natural number. Thus, on neglecting it, we get 5 numbers.
Total required numbers = 180 + 30 + 5 = 215

Page No 16.16:

Question 40:

How many five digit telephone numbers can be constructed using the digits 0 to 9. If each number starts with 67 and no digit appears more than once?

Answer:

The first two digits are fixed as 67.
As repetition of digits in not allowed, the number of available digits to fill the remaining places is 8.
The third place can be filled in 8 ways.
The fourth place can be filled in 7 ways.
The fifth place can be filled in 6 ways.
Total number of such telephone numbers = 8×7×6 = 336

Page No 16.16:

Question 41:

Find the number of ways in which 8 distinct toys can be distributed among 5 childrens.

Answer:

Each of the toy can be distributed in 5 ways.
∴ Total number of ways of distributing the toys = 5×5×5×5×5×5×5×5 = 58

Page No 16.16:

Question 42:

Find the number of ways in which one can post 5 letters in 7 letter boxes.

Answer:

Each of the 5 letters can be posted in any one of the 7 letter boxes.
∴ Required number of ways of posting the letters = 7×7×7×7×7 = 75

Page No 16.16:

Question 43:

Three dice are rolled. Find the number of possible outcomes in which at least one die shows 5.

Answer:

Required number of possible outcomes = (Total number of outcomes - Number of possible outcomes in which 5 does not appear on any of the dice.)
Total number of outcomes when a single dice is rolled = 6
∴ Total number of outcomes when two dice are rolled = 6×6
Similarly, total number of outcomes when three dice are rolled = 6×6×6 = 216
Number of possible outcomes in which 5 does not appear on any dice = 5×5×5 = 125
∴ Required number of possible outcomes = 216 - 125 = 91

Page No 16.16:

Question 44:

Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball.

Answer:

Any one of the twenty balls can be put in the first box. Thus, there are twenty different ways for this.
Now, remaining 19 balls are to be put into the remaining 4 boxes. This can be done in 419 ways because there are four choices for each ball.
∴ Required number of ways = 20×419

Page No 16.16:

Question 45:

In how many ways can 5 different balls be distributed among three boxes?

Answer:

Each ball can be distributed in 3 ways.
∴  Required number of ways in order to distribute 5 balls = 3×3×3×3×3 = 243

Page No 16.16:

Question 46:

In how many ways can 7 letters be posted in 4 letter boxes?

Answer:

Each letter can be posted in any one of the 4 letter boxes.
Number of ways of posting one letter = 4
∴ Required number of ways of posting the 7 letters = 4×4×4×4×4×4×4 = 47

Page No 16.16:

Question 47:

In how many ways can 4 prizes be distributed among 5 students, when
(i) no student gets more than one prize?
(ii) a student may get any number of prizes?
(iii) no student gets all the prizes?

Answer:

(i) Since no student gets more than one prize; the first prize can be given to any one of the five students.
     The second prize can be given to anyone of the remaining 4 students.
     Similarly, the third prize can be given to any one of the remaining 3 students.
    The last prize can be given to any one of the remaining 2 students.
    ∴ Required number of ways = 5×4×3×2 = 5!

(ii) Since a student may get any number of prizes, the first prize can be given to any of the five students. Similarly, the rest of the three prizes can be given to the each of the remaining 4 students.
∴ Required number of ways = 5×5×5×5 = 625

(iii) None of the students gets all the prizes.
  ∴ Required number of ways = {Total ways of distributing the prizes in a condition wherein a student may get any number of prizes - Total ways in a condition in which a student receives all the prizes} =
625 - 5 = 620

Page No 16.16:

Question 48:

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.

Answer:

Total number of lamps = 10

A lamp can be either switched on or switched off.

Since, at least one lamp is to be kept switched on,

∴ The total number of ways are 210-1 = 1023.

Thus, the number of ways in which the hall can be illuminated are 1023.



Page No 16.28:

Question 1:

Evaluate each of the following:
(i) P38

(ii) P410

(iii) P66

(iv) P(6, 4)

Answer:

(i) 8P3
 nPr = n!(n-r)!

8P3  =8!(8-3)!                                     

             = 8!5!

             =8(7)(6)(5!)5!=8×7×6 = 336
 
ii 10P4=10!(10-4)!              = 10!6!             =10(9)(8)(7)(6!)6!             =10×9×8×7              = 5040  


 iii 6P6=6!(6-6)!             = 6!0!             = 6!1             (Since , 0! = 1)            = 720     

(iv) P(6,4)
It can also be written as 6P4 .
6P4 = 6!2!         = 6(5)(4)(3)(2!)2!         =6×5×4×3          = 360

Page No 16.28:

Question 2:

If P (5, r) = P (6, r − 1), find r.

Answer:

P (5, r) = P (6, r − 1)
or  5Pr = 6Pr-1
5!5-r!=6!6-r+1!6-r+1!5-r!=6!5!(7-r)!5-r!=65!5!7-r6-r5-r!5-r!=67-r6-r= 67-r6-r =3×2On comparing the above two equations, we get:7-r=3 r= 4

Page No 16.28:

Question 3:

If 5 P(4, n) = 6. P (5, n − 1), find n.

Answer:

5 P(4, n) = 6. P (5, n − 1)
5 4Pn = 65Pn-1
5×4!4-n!=6×5!5-n+1!5×6-n!4-n!=6×5!4!5×6-n6-n-16-n-2!4-n=6×5×4!4!5×6-n5-n4-n!4-n=6×56-n5-n = 66-n5-n = 3×2On comparing the LHSand the RHS, we get:6-n =3n=3

Page No 16.28:

Question 4:

If P (n, 5) = 20. P(n, 3), find n

Answer:

P (n, 5) = 20. P(n, 3)
n!(n-5)!=20×n!(n-3)!n!n!=20×(n-5)!(n-3)!1 = 20×(n-5)!(n-3)(n-4)(n-5)!(n-3)(n-4) = 20(n-3)(n-4) = 5×4On comparing the two sides, we get:n-3 = 5n=8

Page No 16.28:

Question 5:

If P4n = 360, find the value of n.

Answer:

nP4 = 360
n!n-4! = 360nn-1n-2n-3n-4!n-4!=360nn-1n-2n-3 = 360nn-1n-2n-3 = 6×5×4×3On comparing the two sides, we get:n = 6

Page No 16.28:

Question 6:

If P (9, r) = 3024, find r.

Answer:

P (9, r) = 3024
9!9-r!=30249!9-r!=9×8×7×69!9-r! = 9×8×7×6×5!5!9!9-r!=9!5!9-r!=5! 9-r = 5 r = 4

Page No 16.28:

Question 7:

If P(11, r) = P (12, r − 1) find r.

Answer:

P(11, r) = P (12, r − 1)
11!11-r!=12!13-r!13-r11-r!=12!11!13-r12-r11-r!11-r!=12×11!11!13-r12-r= 1213-r12-r = 4×3On comparing the two sides, we get:13-r = 4r=9

Page No 16.28:

Question 8:

If P (n, 4) = 12 . P (n, 2), find n.

Answer:

P (n, 4) = 12 . P (n, 2)
n!n-4!=12×n!n-2!n-2!n-4!=12×n!n!n-2n-3n-4!n-4!=12n-2n-3 = 12n-2n-3= 4×3On comparing the LHSand the RHS, we get:n-2= 4n=6

Page No 16.28:

Question 9:

If P (n − 1, 3) : P (n, 4) = 1 : 9, find n.

Answer:

P (n − 1, 3):P (n, 4) = 1:9

n-1!(n-1-3)!×(n-4)!(n)!=19n-1!n-4!×n-4!n!=19n-1!n!=19n-1!nn-1!=191n= 19n = 9

Page No 16.28:

Question 10:

If P (2n − 1, n) : P (2n + 1, n − 1) = 22 : 7 find n.

Answer:

P (2n − 1, n):P (2n + 1, n − 1) = 22:7

2n-1!2n-1-n!×2n+1-n+1!2n+1!=2272n-1!n-1!×n+2!2n+1!=2272n-1!n-1!×n+2n+1nn-1!2n+12n2n-1!=227n+2n+1n2n+12n=227n+2n+122n+1= 2277n2+21n+14 = 88n +447n2-67n-30=07n2-70n+3n-30=0n-107n+3=0 n = 10 or -37Since n cannot be negative, it is equal to10.

Page No 16.28:

Question 11:

If P (n, 5) : P (n, 3) = 2 : 1, find n.

Answer:

We have, P (n, 5):P (n, 3) = 2:1

n!n-5!×n-3!n!=21n!n-5!×n-3n-4n-5!n!=21n-3n-4 = 2n-3n-4= 2×1Thus, on comparing the LHSand the RHS in above expression, we get,n-3=2n=5 

Page No 16.28:

Question 12:

Prove that:
1 . P (1, 1) + 2 . P (2, 2) + 3 . P (3, 3) + ... + n . P (n, n) = P (n + 1, n + 1) − 1.

Answer:

1.P (1, 1) + 2. P (2, 2) + 3. P (3, 3) + ... + n . P (n, n) = P (n + 1, n + 1) − 1
P (n,n) = n!
1.1! + 2.2! + 3.3! ......+ n.n! = (n+1)! − 1
LHS = 1.1! + 2.2! + 3.3! ......+ n.n!
       =r=1nr.r!=r=1nr+1-1 r!= r=1nr+1 r! - r! =  r=1n{(r+1)!-r!} = 2!-1!+3!-2!+...n+1!-n!=n+1!-1! = n+1!-1 = RHS  Hence, proved.

Page No 16.28:

Question 13:

If P (15, r − 1) : P (16, r − 2) = 3 : 4, find r.

Answer:

P (15, r − 1):P (16, r − 2) = 3:4

15!15-r+1!×(16-r+2)!16!=3415!16-r!×18-r!16×15!=3418-r17-r16-r!16-r!16=3418-r17-r = 1218-r17-r = 4×3On comparing the LHSand the RHSin above expression, we get:18-r= 14r=14

Page No 16.28:

Question 14:

If n +5Pn +1 = 11 (n-1)2n +3Pn, find n.

Answer:

n +5Pn +1 = 11(n-1)2n +3Pn
n+5!n+5-n-1!=11n-12×n+3!n+3-n!n+5!4!=11n-12×n+3!3!n+5!n+3!=11n-12×4!3!n+5n+4n+3!n+3!=11n-12×4×3!3!n+5n+4 = 22n-1n2+9n+20 = 22n-22n2-13n+42=0n = 7,6

Page No 16.28:

Question 15:

In how many ways can five children stand in a queue?

Answer:

Required number of ways = Number of arrangements of all the children = 5P5 = 5!
We know:
nPn = n!
∴ 5P5  = 5! = 120  

Page No 16.28:

Question 16:

From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?

Answer:

Here, we need to permute 2 teachers out of the 36 available teachers.
It can also be understood as the arrangement of 36 teachers, taken two at a time.
∴ Required number of ways = 36P2
=36!36-2!=36!34!=36×35×34!34!=36×35 = 1260

Page No 16.28:

Question 17:

Four letters E, K, S and V, one in each, were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them?

Answer:

Here, we need to find out the number of pairs of the letters that can be formed with the 4 letters.
Required number of ordered pairs = Number of arrangements of  four letters, taken two at a time = 4P2
=4!4-2!=4!2!=4×3×2!2! = 4×3=12

Page No 16.28:

Question 18:

Four books, one each in Chemistry, Physics, Biology and Mathematics, are to be arranged in a shelf. In how many ways can this be done?

Answer:

Here, all the four books are to be arranged on a shelf. This means that we have to find the number of arrangements of the books, taken all at a time.
⇒ 4P4
Now, nPn = n!
Similarly, 4P4  = 4! = 24

Page No 16.28:

Question 19:

Find the number of different 4-letter words, with or without meanings, that can be formed from the letters of the word 'NUMBER'.

Answer:

Here, we need to permute four of the letters from the available 6 letters of the word NUMBER.
Number of different four letter words = Number of arrangements of 6 letters, taken 4 at a time =6 P4
=6!(6-4)!=6!2!=6×5×4×3×2!2!=6×5×4×3 = 360

Page No 16.28:

Question 20:

How many three-digit numbers are there, with distinct digits, with each digit odd?

Answer:

The odd digits are 1, 3, 5, 7 and 9.

Required number of ways = Number of arrangements of five digits ( 1, 3, 5, 7 and 9), taken three at a time = 5P3
=5!(5-3)!=5!2!=5×4×3×2!2!=5×4×3=60

Page No 16.28:

Question 21:

How many words, with or without meaning, can be formed by using all the letters of the word 'DELHI', using each letter exactly once?

Answer:

There are 5 letters in the word DELHI.
Number of 5 letter words = Number of arrangements of 5 letters, taken 5 at a time
= 5P5 = 120

Page No 16.28:

Question 22:

How many words, with or without meaning, can be formed by using the letters of the word 'TRIANGLE'?

Answer:

There are 8 letters in the word TRIANGLE.
∴ Number of 8 letter words = Number of arrangements of 8 letters, taken 8 at a time
= 8P8 = 8!



Page No 16.29:

Question 23:

There are two works each of 3 volumes and two works each of 2 volumes; In how many ways can the 10 books be placed on a shelf so that the volumes of the same work are not separated?

Answer:

There are 4 different types of works.
∴ Number of arrangements of these 4 works, taken 4 at a time = 4!
Of these 4 works, two of the works with 3 volumes each can be arranged in 3! ways each and two of the works with 2 volumes each can be arranged in 2! ways.
Total number of arrangements  = 4!×(3!×3!) ×(2!×2!) = 3456

Page No 16.29:

Question 24:

There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible, correct or incorrect, answers are there to this question?

Answer:

Each answer is an arrangement of the 6 items of column B keeping the order of column A fixed.
∴ Total number of answers = Number of arrangements of items in column B = 6P6 = 6! = 720

Page No 16.29:

Question 25:

How many three-digit numbers are there, with no digit repeated?

Answer:

Total number of 3-digit numbers = Number of arrangements of 10 numbers, taken 3 at a time = 10P3 = 10!7!=10×9×8 = 720
Total number of 3-digit numbers, having 0 at its hundred's place = 9P2 = 9!7!=9×8 = 72
Total number of 3-digit numbers with distinct digits = 10P3 - 9P2 = 720 - 72 = 648

Page No 16.29:

Question 26:

How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?

Answer:

Total available digits = 10
Out of these, 3 and 5 have already been used to make the first two digits.
∴ Number of available digits = 8
The telephone number consists of 6 digits. The initial numbers have already been fixed as 35.

Since repetition is not allowed, the number of telephone numbers that can be formed is equal to the number of  arrangements of the 8 digits, taken 4 at a time.
⇒ 8P4 = 8!4!=8×7×6×5 = 1680

Page No 16.29:

Question 27:

In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?

Answer:

Number of arrangements of the boys = Number of arrangements of the 6 boys taken 6 at a time = 6!
Number of arrangements of the girls = Number of arrangements of the 5 girls taken 5 at a time = 5!
Total number of arrangements = 6!×5! = 86400

Page No 16.29:

Question 28:

If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of  permutations of x − 11 things taken all at a time such that a = 182 bc, find the value of x.

Answer:

a = x+2Px+2 = (x+2)!
b  = xP11 = x!(x-11)!
c = x-11Px-11(x-11)!

            a = 182 bc
      
(x+2)! = 182 x!x-11! ×x-11!
      (x+2)! = 182 (x!)

       x+2!x! = 182 x+2x+1= 182x+2x+1= 14×13  x+2 = 14x=12    
        

Page No 16.29:

Question 29:

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Answer:

Total number of arrangements of 9 digits, taken 3 at a time = 9P3
∴ Total 3-digit numbers that can be formed by using the digits 1 to 9, if no digit is repeated = 9P3 = 9×8×7 = 504  

Page No 16.29:

Question 30:

How many 3-digit even number can be made using the digits 1, 2, 3, 4, 5, 6, 7, if no digits is repeated?

Answer:

In order to find the number of even digits, we fix the unit's digit as an even digit.

Fixing the unit's digit as 2:
Number of arrangements possible = 6P2  = 6×5 = 30

Similarly, fixing the unit's digit as 4:
Number of arrangements possible = 6P2  = 6×5 = 30

Fixing the unit's digit as 6:
Number of arrangements possible = 6P2  = 6×5 = 30
∴ Number of 3-digit even numbers that can be formed = 30 + 30 + 30 = 90

Page No 16.29:

Question 31:

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated? How many of these will be even?

Answer:

Number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 = Number of arrangements of 5 digits taken 4 at a time = 5P4 = 5! = 120
Now, these numbers also consist of numbers in which the last digit is an odd digit.
So, in order to find the number of even digits, we subtract the cases in which the unit's digit have been fixed as an odd digit.
Fixing the unit's digit as 1:
Number of arrangements possible = 4P3  = 4!
Fixing the unit's digit as 3:
Number of arrangements possible = 4P= 4!
Fixing the unit's digit as 5:
Number of arrangements possible = 4P3  = 4!
∴ Number of 4-digit even numbers that can be formed = 120 -4!-4!-4! = 120-24-24-24 = 48

Page No 16.29:

Question 32:

All the letters of the word 'EAMCOT' are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

Answer:

We note that, there are 3 consonants M, C, T and 3 vowels E, A, O.

Since, no two vowels have to be together, the possible choice for volwels are the blank spaces,
_M_C_T_

These vowels can be arranged in 4P3 ways.
3 consonants can be arranged in 3! ways.

Hence, the required numbers of ways = 3! × 4P3 = 144 ways.



Page No 16.36:

Question 1:

In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?

Answer:

There are 7 letters in the word FAILURE.
We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.
There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.

By fundamental principle of counting:
Total number of arrangements = 4!×4! = 576

Page No 16.36:

Question 2:

In how many ways can the letters of the word 'STRANGE' be arranged so that
(i) the vowels come together?
(ii) the vowels never come together? and
(iii) the vowels occupy only the odd places?

Answer:

(i)  Number of vowels = 2
Number of consonants = 5
Considering the two vowels as a single entity, we are now to arrange 6 entities taken all at a time.
Total number of ways = 6!
Also, the two vowels can be mutually arranged amongst themselves in 2! ways.

By fundamental principle of counting:
Total number of words that can be formed = 6!×2! = 1440

(ii) Total number of words that can be made with the letters of the word STRANGE = 7! = 5040
Number of words in which vowels always come together = 1440
∴ Number of words in which vowels do not come together = 5040 -1440 = 3600

(iii) There are 7 letters in the word STRANGE.
We wish to find the total number of arrangements of these 7 letters so that the vowels occupy only odd positions.
There are 2 vowels and 4 odd positions.
These 2 vowels can be arranged in the 4 positions in 4×3 ways, i.e. 12 ways.
The remaining 5 consonants can be arranged in the remaining 5 positions in 5! ways.
By fundamental principle of counting:
Total number of arrangements = 12×5! = 1440

Page No 16.36:

Question 3:

How many words can be formed from the letters of the word 'SUNDAY'? How many of these begin with D?

Answer:

Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720
Fixing the first letter as D:
Number of arrangements of the remaining 5 letters, taken 5 at a time = 5! = 120
Number of words with the starting letter D = 120



Page No 16.37:

Question 4:

How many words can be formed out of the letters of the word, 'ORIENTAL', so that the vowels always occupy the odd places?

Answer:

There are 8 letters in the word ORIENTAL.
We wish to find the total number of arrangements of these 8 letters so that the vowels occupy only odd positions.
There are 4 vowels and 4 odd positions.
These 4 vowels can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 consonants can be arranged in the remaining 4 positions in 4! ways.
By fundamental principle of counting:
Total number of arrangements = 4!×4! = 576

Page No 16.37:

Question 5:

How many different words can be formed with the letters of word 'SUNDAY'? How many of the words begin with N? How many begin with N and end in Y?

Answer:

Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720
Now, if we fix the first letter as N, the remaining 5 places can be filled with the remaining 5 letters in 5! ways, i.e. 120.
If we fix the first letter as N and the last word as Y:
Remaining 4 places can be filled with 4 letters in 4! ways = 24

Page No 16.37:

Question 6:

How many different words can be formed from the letters of the word 'GANESHPURI'? In how many of these words:
(i) the letter G always occupies the first place?
(ii) the letters P and I respectively occupy first and last place?
(iii) the vowels are always together?
(iv) the vowels always occupy even places?

Answer:

The word GANESHPURI consists of 10 distinct letters.
Number of letters = 10!

(i) If we fix the first letter as G, the remaining 9 letters can be arranged in 9! ways to form the words.
∴ Number of words starting with the letter G = 9!

(ii) If we fix the first letter as P and the last letter as I, the remaining 8 letters can be arranged in 8! ways to form the words.
∴  Number of words that start with P and end with I = 8!

(iii) The word GANESHPURI consists of 4 vowels. If we keep all the vowels together, we have to consider them as a single entity.
So, we are left with the remaining 6 consonants and all the vowels that are taken together as a single entity. This gives us a total of 7 entities that can be arranged in 7! ways.
Also, the 4 vowels can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:
Total number of arrangements = 7!×4! words.

(iv) The word GANESHPURI consists of 4 vowels that have to be arranged in the 5 even places. This can be done in 5! ways.
Now, the remaining 6 consonants can be arranged in the remaining 6 places in 6! ways.
Total number of words in which the vowels occupy even places = 5!×6!

Page No 16.37:

Question 7:

How many permutations can be formed by the letters of the word, 'VOWELS', when
(i) there is no restriction on letters?
(ii) each word begins with E?
(iii) each word begins with O and ends with L?
(iv) all vowels come together?
(v) all consonants come together?

Answer:

(i) The word VOWELS consists of 6 distinct letters that can be arranged amongst themselves in 6! ways.
∴ Number of words that can be formed with the letters of the word VOWELS, without any restriction = 6! = 720

(ii) If we fix the first letter as E, the remaining 5 letters can be arranged in 5! ways to form the words.
∴ Number of words starting with the E = 5! = 120

(iii) If we fix the first letter as O and the last letter as L, the remaining 4 letters can be arranged in 4! ways to form the words.
∴  Number of words that start with O and end with L = 4! = 24

(iv) The word VOWELS consists of 2 vowels.
If we keep all the vowels together, we have to consider them as a single entity.
Now, we are left with the 4 consonants and all the vowels that are taken together as a single entity.
This gives us a total of 5 entities that can be arranged in 5! ways.
It is also to be considered that the 2 vowels can be arranged in 2! ways amongst themselves.

By fundamental principle of counting:
∴ Total number of arrangements = 5!×2! = 240

(v) The word VOWELS consists of 4 consonants.
If we keep all the consonants together, we have to consider them as a single entity.
Now, we are left with the 2 vowels and all the consonants that are taken together as a single entity.
This gives us a total of 3 entities that can be arranged in 3! ways.
It is also to be considered that the 4 consonants can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:
∴ Total number of arrangements = 3!×4! = 144

Page No 16.37:

Question 8:

How many words can be formed out of the letters of the word 'ARTICLE', so that vowels occupy even places?

Answer:

The word ARTICLE consists of 3 vowels, which have to be arranged in 3 even places. This can be done in 3! ways.
Now, the remaining 4 consonants can be arranged in the remaining 4 places in 4! ways.
∴ Total number of words in which the vowels occupy only even places = 3!×4!  = 144

Page No 16.37:

Question 9:

In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?

Answer:

We arrange any 2 men in 7P2 ways and then the wives of the remaining 5 men can be arranged in 5P2 ways. This is because these two men should not be with their respective wives.
∴ By fundamental principle of counting, the required number of ways  = 7P2 ×5P2
=7!5!×5!3! = 7!3!  = 840

Page No 16.37:

Question 10:

m men and n women are to be seated in a row so that no two women sit together. if m > n then show that the number of ways in which they can be seated as
m! (m+1)!(m-n+1) !

Answer:

'm' men can be seated in a row in m! ways.
'm' men will generate (m+1) gaps that are to be filled by 'n' women = Number of arrangements of (m+1) gaps, taken 'n' at a time = m+1Pn = m+1!m+1-n!
∴ By fundamental principle of counting, total number of ways in which they can be arranged = m!m+1!m-n+1!

Page No 16.37:

Question 11:

How many words (with or without dictionary meaning) can be made from the letters in the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time?
(ii) all letters are used at a time.
(iii) all letters are used but first is vowel.

Answer:

(i) The word MONDAY consists of 6 letters.
Number of words formed using 4 letters = Number of arrangements of 6 letters, taken 4 at a time = 6P4 = 6!2!= 6×5×4×3 = 360
(ii) Number of words formed using all the letters = Number of arrangements of 6 letters, taken all at a time = 6P6 = 6! = 720
(iii) The word MONDAY consists of 2 vowels and 4 consonants.
The first letter has to be a vowel, which is to be chosen from the two vowels.
This can be done in two ways. The remaining 5 letters can be arranged in 5! ways to form 6 letter words.
⇒ 2 ×5! = 240

Page No 16.37:

Question 12:

How many three letter words can be made using the letters of the word 'ORIENTAL'?

Answer:

The word ORIENTAL consists of 8 letters. In order to make three letter words, we need to permute these 8 letters, taken three at a time.
⇒ 8P3 = 8×7×6 = 336



Page No 16.4:

Question 1:

Compute:
(i) 30!28!

(ii) 11!-10!9!

(iii) L.C.M. (6!, 7!, 8!)

Answer:


(i) 30!28! = 30×29×28!28!     n!= n(n-1)!               =30×29               = 870

(ii) 11!-10!9!=11×10×9! - 10×9!9!           n! =n(n-1)!                        =9!(110-10)9!                        =100

(iii) LCM of (6!,7! and 8!):
 n! = n(n-1)!
Therefore, (6!,7! and 8!) can be rewritten as:
8! = 8×7×6!
7! = 7×6!
6! = 6!
∴ LCM of (6!,7! and 8!) = LCM [8×7×6!, 7×6! ,  6!]  = 8×7×6! = 8!

Page No 16.4:

Question 2:

Prove that 19!+110!+111!=12211!

Answer:

LHS=19!+110!+111!        =19!+110×9!+111×10×9!        =110+11+111×10×9!        =12211! = RHSHence, proved.

Page No 16.4:

Question 3:

Find x in each of the following:
(i) 14!+15!=x6!

(ii) x10!=18!+19!

(iii) 16!+17!=x8!

Answer:

(i) 14!+15! = x6!14!+15(4!)= x6!5+15(4!) = x6! 65!= x6! 65!= x6×5!x = 36

 (ii) x10! =18!+19!x10! =18!+19(8!)x10! =9+19(8!) x10! = 109!x10×9! = 109!x=100

 (iii) 16!+17!=x8!16!+17(6!)=x8!7+17(6!)=x8! 87!=x8! 87!=x8×7!x = 64

Page No 16.4:

Question 4:

Convert the following products into factorials:
(i) 5 · 6 · 7 · 8 · 9 · 10
(ii) 3 · 6 · 9 · 12 · 15 · 18
(iii) (n + 1) (n + 2) (n + 3) ... (2n)
(iv) 1 · 3 · 5 · 7 · 9 ... (2n − 1)

Answer:

i 5×6×7×8×9×10=1×2×3×4×5×6×7×8×9×101×2×3×4                                   =10!4!
    
ii 3×6×9×12×15×18=3×1×3×2×3×3×3×4×3×5×3×6                                   =361×2×3×4×5×6                                   =366!

iii n+1n+2n+3...2n=123...nn+1n+2n+3...2n123...n                                                  =2nn!

 iv 135.........2n-1  =135.........2n-1246.........2n246.........2n                                                 =12345.........2n-12n2n123.........n                                                =2n!2nn!
    
     

Page No 16.4:

Question 5:

Which of the following are true:
(i) (2 +3)! = 2! + 3!
(ii) (2 × 3)! = 2! × 3!

Answer:

(i)  LHS = (2 +3)!
              = 5!
              = 120
    RHS = 2! + 3!
             = 2 + 6
             = 8
   Since  LHS ≠ RHS,
      Thus, (i) is false.

(ii) LHS =  (2 × 3)!
              = 6!
              = 720
     RHS = 2! × 3!
              = 2 × 6
              = 12
     LHS ≠ RHS
    Thus, (ii) is false.

Page No 16.4:

Question 6:

Prove that: n! (n + 2) = n! + (n + 1)!

Answer:

RHS = n! + (n + 1)!
        =  n! + (n + 1)(n!)
        = n! ( 1+ n + 1)
        = n! (n+2) = LHS
Hence, proved.

Page No 16.4:

Question 7:

If (n + 2)! = 60 [(n − 1)!], find n.

Answer:

(n + 2)! = 60 [(n − 1)!]
(n + 2)×(n + 1)×(n)×( 1)! = 60 [(n − 1)!]
(n + 2)×(n + 1)×(n) = 60
(n + 2)×(n + 1)×(n) = 5×4×3
n = 3

Page No 16.4:

Question 8:

If (n + 1)! = 90 [(n − 1)!], find n.

Answer:

(n + 1)! = 90 [(n − 1)!]
(n + 1)×(n)×(n−1)! = 90[(n − 1)!]
(n + 1)×(n) = 90
(n + 1)×(n) = 10 × 9
On comparing, we get:
n = 9



Page No 16.42:

Question 1:

Find the number of words formed by permuting all the letters of the following words:
(i) INDEPENDENCE
(ii) INTERMEDIATE
(iii) ARRANGE
(iv) INDIA
(v) PAKISTAN
(vi) RUSSIA
(vii) SERIES
(viii) EXERCISES
(ix) CONSTANTINOPLE

Answer:

(i) This word consists of 12 letters that include three Ns, two Ds and four Es.
The total number of words is the number of arrangements of 12 things, of which 3 are similar to one kind, 2 are similar to the second kind and 4 are similar to the third kind.
⇒ 12!3!2!4! = 1663200

(ii) This word consists of 12 letters that include two Is, two Ts and three Es.
The total number of words is the number of arrangements of 12 things, of which 2 are similar to one kind, 2 are similar to the second kind and 3 are similar to the third kind.
12!2!2!3! = 19958400

(iii) This word consists of 7 letters that include two Rs, and two As.
The total number of words is the number of arrangements of 7 things, of which 2 are similar to one kind and 2 are similar to the second kind.
7!2!2! = 1260

(iv) This word consists of 5 letters that include two Is.
The total number of words is the number of arrangements of 5 things, of which 2 are similar to one kind.
5!2! = 60

(v) This word consists of 8 letters that include two As.
The total number of words is the number of arrangements of 7 things, of which 2 are similar to one kind.
8!2! = 20160

(vi) This word consists of 6 letters that include two Ss.
The total number of words is the number of arrangements of 6 things, of which 2 are similar to one kind.
6!2! = 360

(vii) This word consists of 6 letters that include two Ss and two Es.
The total number of words is the number of arrangements of 6 things, of which 2 are similar to one kind and 2 are similar to the second kind.
6!2!2! = 180

(viii) This word consists of 9 letters that include three Es and two Ss.
The total number of words is the number of arrangements of 9 things, of which 2 are similar to one kind and 2 are similar to the second kind.
9!2!3! = 30240

(ix) This word consists of 14 letters that include three Ns, two Os and two Ts.
The total number of words is the number of arrangements of 14 things, of which 3 are similar to one kind, 2 are similar to the second kind and 2 are similar to the third kind.
14!3!2!2! =14!24

Page No 16.42:

Question 2:

In how many ways can the letters of the word 'ALGEBRA' be arranged without changing the relative order of the vowels and consonants?

Answer:

The relative positions of all the vowels and consonants is fixed.
The first letter is a vowel. It can be selected out of the 3 three vowels, of which two are same. So, the vowels can be arranged in selecting 3 things, of which two are of the same kind
⇒ 3!2!
The second, third, fifth and sixth letters are consonants that can be filled by the available 4 consonants in 4! ways.
∴ By fundamental principle of counting, the number of words that can be formed = 4!×3!2! = 72

Page No 16.42:

Question 3:

How many words can be formed with the letters of the word 'UNIVERSITY', the vowels remaining together?

Answer:

The word UNIVERSITY consists of 10 letters that include four vowels of which two are same.
Thus, the vowels  can be arranged amongst themselves in 4!2!ways.
Keeping the vowels as a single entity, we are left with 7 letters, which can be arranged in 7! ways.
By fundamental principle of counting, we get,
Number of words =  7!× 4!2! = 60480

Page No 16.42:

Question 4:

Find the total number of arrangements of the letters in the expression a3 b2 c4 when written at full length.

Answer:

When expanded, a3 b2 c4   would result in total 9 letters.
This is same as permuting 9 things, of which 3 are similar to the first kind, 2 are similar to the second kind and four are similar to the third kind, i.e. three as , two bs and four cs.
Required number of arrangements = 9!3!2!4! = 1260



Page No 16.43:

Question 5:

How many words can be formed with the letters of the word 'PARALLEL' so that all L's do not come together?

Answer:

The word PARALLEL consists of 8 letters that include two As and three Ls.
Total number of words that can be formed using the letters of the word PARALLEL = 8!2!3! = 3360
Number of words in which all the Ls come together is equal to the condition if all three Ls are considered as a single entity.
So, we are left with total 6 letters that can be arranged in 6!2! ways (divided by 2! since there are two As), which is equal to 360.
Number of words in which all Ls do not come together = Total number of words - Number of words in which all the Ls come together
                                                                                        =  3360 -360
                                                                                        = 3000

Page No 16.43:

Question 6:

How many words can be formed by arranging the letters of the word 'MUMBAI' so that all M's come together?

Answer:

The word MUMBAI consists of 6 letters taht include two Ms.
When we consider both the Ms as a single entity, we are left with 5 entities that can be arranged in 5! ways.
Total number of words that can be formed with all the Ms together = 5! = 120

Page No 16.43:

Question 7:

How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Answer:

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in 4!2!2!ways.
The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in 3!2!ways.
By fundamental principle of counting:
Required number of arrangements = 4!2!2!×3!2! = 18

Page No 16.43:

Question 8:

How many different signals can be made from 4 red, 2 white and 3 green flags by arranging all of them vertically on a flagstaff?

Answer:

We have to arrange 9 flags, out of which 4 are of one kind (red), 2 are of another kind (white) and 3 are of the third kind (green).
∴ Total number of signals that can be generated with these flags = 9!4!2!3! = 1260

Page No 16.43:

Question 9:

How many number of four digits can be formed with the digits 1, 3, 3, 0?

Answer:

The given digits are 1, 3, 3, 0.
Total numbers that can be formed with these digits = 4!2!
Now, these numbers also include the numbers in which the thousand's place is 0.
But, to form a four digit number, this is not possible.
∴ Numbers in which the thousand's place is fixed as zero = Ways of arranging the remaining digits (1,3 and 3) in three places = 3!2!
∴ Four digit numbers = Total numbers - Numbers in which the thousand's place is 0
                                 =  4!2!- 3!2! = 9

Page No 16.43:

Question 10:

In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together?

Answer:

The word ARRANGE consists of 7 letters including two Rs and two As, which can be arranged in 7!2!2!ways.
∴ Total number of words that can be formed using the letters of the word ARRANGE = 1260
Number of words in which the two Rs are always together = Considering both Rs as a single entity
                                                                                               = Arrangements of  6 things of which two are same (two As)
                                                                                               = 6!2!
                                                                                               = 360
Number of words in which the two Rs are never together = Total number of words - Number of words in which the two Rs are always together
                                                                                           = 1260-360
                                                                                           = 900

Page No 16.43:

Question 11:

How many different numbers, greater than 50000 can be formed with the digits 0, 1, 1, 5, 9.

Answer:

Numbers greater than 50000 can either have 5 or 9 in the first place and will consist of 5 digits.
Number of arrangements having 5 as the first digit = 4!2!

Number of arrangement having 9 as the first digit = 4!2!
∴ Required arrangements = 4!2!+4!2! = 24

Page No 16.43:

Question 12:

How many words can be formed from the letters of the word 'SERIES' which start with S and end with S?

Answer:

The word SERIES consists of 6 letters including two Ss and two Es.
The first and the last letters are fixed as S.
Now, the remaining four letters can be arranged in 4!2! ways = 12

Page No 16.43:

Question 13:

How many permutations of the letters of the word 'MADHUBANI' do not begin with M but end with I?

Answer:

Number of words that only end with I = Number of permutations of the remaining 8 letters, taken all at a time = 8!2!

Number of words that start with M and end with I = Permutations of the remaining 7 letters, taken all at a time =7!2!
Number of words that do not begin with M but end with I = Number of words that only end with I - Number of words that start with M and end with I
                                                                                         = 8!2! - 7!2! 
                                                                                          = 17640

Page No 16.43:

Question 14:

Find the number of numbers, greater than a million, that can be formed with the digits 2, 3, 0, 3, 4, 2, 3.

Answer:

One million (1,000,000) consists of 7 digits.
We have digits 2, 3, 0, 3, 4, 2 and 3.
Numbers formed by arranging all these seven digits = 7!2!3!
But, these numbers also include the numbers whose first digit is 0.
This is invalid as in that case the number would be less than a million.
Total numbers in which the first digit is fixed as 0 = Permutations of the remaining 6 digits = 6!2!3!
Numbers that are greater than 1 million = 7!2!3!-6!2!3! = 360

Page No 16.43:

Question 15:

There are three copies each of 4 different books. In how many ways can they be arranged in a shelf?

Answer:

Total number of books = 12
∴ Required number of arrangements = Arrangements of 12 things of which each of the 4 different books has three copies = 12!3!3!3!3! = 12!(3!)4

Page No 16.43:

Question 16:

How many different arrangements can be made by using all the letters in the word 'MATHEMATICS'. How many of them begin with C? How many of them begin with T?

Answer:

The word MATHEMATICS consists of 11 letters that include two Ms, two As, and two Ts.
Total number of arrangements of the letters of the word MATHEMATICS = 11!2!2!2!
Number of words in which the first word is fixed as C = Number of arrangements of the remaining 10 letters, of which there are two As, two Ms and two Ts = 10!2!2!2!
Number of words in which the first word is fixed as T = Number of arrangements of the remaining 10 letters, of which there are two As and two Ms = 10!2!2!

Page No 16.43:

Question 17:

A biologist studying the genetic code is interested to know the number of possible arrangements of 12 molecules in a chain. The chain contains 4 different molecules represented by the initials A (for Adenine), C (for Cytosine), G (for Guanine) and T (for Thymine) and 3 molecules of each kind. How many different such arrangements are possible?

Answer:

Number of molecules in a chain = 12
Number of molecules with initials A = 3
Number of molecules with initials C = 3
Number of molecules with initials G = 3
Number of molecules with initials T = 3
Thus, total arrangements of all the molecules in the chain = Number of arrangements of 12 things of which 3 are similar to the first kind, 3 are similar to the second kind, 3 are similar to the third kind and 3 are similar to the fourth kind
12!3!3!3!3! = 369600

Page No 16.43:

Question 18:

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?

Answer:

Number of red discs = 4
Number of yellow discs = 3
Number of green discs = 2
Total number of discs = 9
Total number of arrangements = Number of arrangements of 9 things of which 4 are similar to the first kind, 3 are similar to the second kind and 2 are similar to the third kind = 9!4!3!2! = 1260

Page No 16.43:

Question 19:

How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Answer:

Numbers greater than a million can be formed when the first digit can be any one out of the given digits 1, 2, 0, 2, 4, 2, 4, except 0.
Number of arrangements of the given digits 1, 2, 0, 2, 4, 2, 4 = Arrangements of 7 things of which 3 are similar to the first kind, and 2 are similar to the second kind = 7!2!3!
But, these arrangements also include the numbers in which the first digit is zero. This will make the number less than a million. So, it needs to be subtracted.
Number where the first digit is zero = Number of arrangements of the remaining 6 digits 1, 2, 2, 4, 2, 4
 = 6!2!3!
Numbers greater than 1 million = 7!2!3!-6!2!3! = 360

Page No 16.43:

Question 20:

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

Answer:

The word ASSASSINATION consists of 13 letters including three As, four Ss, two Ns and two Is.
Considering all the Ss are together or as a single letter, we are left with 10 letters. Out of these, there are three As, two Ns and two Is.
Number of words in which all the Ss are together = Permutations of 10 letters of which three are similar to the first kind, two are similar to the second kind and two are similar to the third kind = 10!2!2!3! = 151200

Page No 16.43:

Question 21:

Find the total number of permutations of the letters of the word 'INSTITUTE'.

Answer:

The word 'INSTITUTE' consists of 9 letters including two Is and three Ts.
Total number of words that can be formed of the word INSTITUTE = Number of arrangements of 9 things of which 2 are similar to the first kind and 3 are similar to the second kind = 9!2!3!

Page No 16.43:

Question 22:

The letters of the word 'SURITI' are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word 'SURITI'.

Answer:

In a dictionary, the words are arranged in the alphabetical order. Thus, in the given problem, we must consider the words beginning with I, I, R, S, T and U.
I will occur at the first place as often as the ways of arranging the remaining 5 letters, when taken all at a time.
Thus, I will occur 5! times.
Similarly, R will occur at the first place the same number of times.
∴ Number of words starting with  I = 5! 
Number of words starting with  R = 5!2!
The word will now start with S, which is as per the requirement of the word SURITI.
Alphabetically, the next letter would be I, i.e. SI. The remaining four letters can be arranged in 4! ways.
Alphabetically, the next letter would now be R, i.e. SR. The remaining four letters can be arranged in 4!2! ways.

Alphabetically, the next letter would now be T, i.e. ST. The remaining four letters can be arranged in 4!2! ways.

Alphabetically, the next letter would now be U, i.e. SU, which is as per the requirement of the word SURITI.
After SU, alphabetically, the third letter would be I, i.e. SUI. Thus, the remaining 3 letters can be arranged in 3! ways.
The next third letter that can come is R, i.e. SUR, which is as per the requirement of the word SURITI.
After SUR, the next letter that will come is I, i.e. SURI, which is as per the requirement of the word SURITI.
The next word arranged in the dictionary will be SURIIT.
Then, the next word will be SURITI.
Rank of the word SURITI in the dictionary = 5! + 5!2! + 4! + 4!2! + 4!2! + 3! + 2 = 236

Page No 16.43:

Question 23:

If the letters of the word 'LATE' be permuted and the words so formed be arranged as in a dictionary, find the rank of the word LATE.

Answer:

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word LATE.
For finding the number of words starting with A, we have to find the number of arrangements of the remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with E, we have to find the number of arrangements of the remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with L, the next alphabetical letter would be A, followed by E and then T, i.e. LAET.
The next alphabetical word would be LATE.
Number of words after which we reach the word LATE = 3!+3!+1+1 = 14

Page No 16.43:

Question 24:

If the letters of the word 'MOTHER' are written in all possible orders and these words are written out as in a dictionary, find the rank of the word 'MOTHER'.

Answer:

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word MOTHER.
For finding the number of words starting with E, we have to find the number of arrangements of the remaining 5 letters.
Number of such arrangements = 5!
For finding the number of words starting with H, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 5!

For finding the number of words starting with M, fixing the next letter as E, we have to find the number of arrangements of the remaining 4 letters, which is 4!.
For finding the number of words starting with M, fixing the next letter as H, we have to find the number of arrangements of the remaining 4 letters, which is 4!.

For finding the number of words starting with M, fixing the second letter as O, and the third letter as E, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, and the third letter as H, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, and the third letter as R, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, the third letter as T, and the fourth letter as E, we have to find the number of arrangements of the remaining 2 letters, which is 2!.

Now, the next word formed would be MOTHER.
Number of words after which we reach the word MOTHER = 5!+5!+4!+4!+3!+3!+3!+2!+1 = 309

Page No 16.43:

Question 25:

If the permutations of a, b, c, d, e taken all together be written down in alphabetical order as in dictionary and numbered, find the rank of the permutation debac.

Answer:

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word 'debac'.
For finding the number of words starting with a, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with b, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with c, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with d, fixing the next letter as a, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as b, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as c, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as e:
First word- deabc
Second word- deacb
Third word- debac

Number of words after which we reach the word debac = 4!+4!+4!+3!+3!+3!+1+1+1 = 93

Page No 16.43:

Question 26:

Find the total number of ways in which six '+' and four '−' signs can be arranged in a line such that no two '−' signs occur together.

Answer:

Six '+' signs can be arranged in a row in 6!6! = 1 way
Now, we are left with seven places in which four different things can be arranged in 7P4 ways.
Since all the four '-' signs are identical, four '-' signs can be arranged in P474! ways, i.e. 35 ways.
Number of ways = 1× 35 = 35



Page No 16.44:

Question 27:

In how many ways can the letters of the word "INTERMEDIATE" be arranged so that:
(i) the vowels always occupy even places?
(ii) the relative order of vowels and consonants do not alter?

Answer:

The word INTERMEDIATE consists of 12 letters that include two Is, two Ts and three Es.
(i) There are 6 vowels (I, I, E, E, E and A) that are to be arranged in six even places = 6!2!3!= 60
The remaining 6 consonants can be arranged amongst themselves in 6!2! ways, which is equal to 360.
By fundamental principle of counting, the number of words that can be formed = 60×360 = 21600

(ii) The relative positions of all the vowels and consonants is fixed.
     Arranging the six vowels at their places, without disturbing their respective places, we can arrange the six vowels in 6!2!3! ways.

Similarly, arranging the remaining 6 consonants at their places, without disturbing their respective places, we can arrange the 6 consonants in 6!2! ways.
By fundamental principle of counting, the number of words that can be formed = 6!2!3!×6!2! = 21600

Page No 16.44:

Question 28:

The letters of the word 'ZENITH' are written in all possible orders. How many words are possible if all these words are written out as in a dictionary? What is the rank of the word 'ZENITH'?

Answer:

In a dictionary, the words are arranged in alphabetical order. Therefore, in the given problem, we must consider the words beginning with E, H, I, N, T and Z.

∴ Number of words starting with  E = 5! = 120
    Number of words starting with  H = 5! = 120
    Number of words starting with  I = 5! = 120
    Number of words starting with  N = 5! = 120
    Number of words starting with  T = 5! = 120
Now, the word will start with the letter Z.
After Z, alphabetically, the next letter would be E, which is as per the requirement of the word ZENITH.
After ZE, alphabetically, the next letter would be H, i.e. ZEH. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be I, i.e. ZEI. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be N, which as per the requirement of the word ZENITH.

After ZEN, alphabetically, the next letter would be H, i.e. ZENH. The remaining two letters can be arranged in 2! ways.
The next letter would now be I, i.e. ZENI, which is as per the requirement of the word ZENITH.
H will come after ZENI, which would be followed by T.
The word formed is ZENIHT.
The next word would be ZENITH.
 Total number of intermediate words = 5×120 + 3! + 3! + 2! + 1 + 1 = 616



Page No 16.45:

Question 1:

In how many ways can 4 letters be posted in 5 letter boxes?

Answer:

Each of the letter can be posted in anyone of the letter boxes.
This means that every letter can be posted in 5 ways.
∴ Total number of ways of posting 4 letters = 5×5×5×5 = 54

Page No 16.45:

Question 2:

Write the number of 5 digit numbers that can be formed using digits 0, 1 and 2.

Answer:

Number of ways in which the first digit can be filled = Number of digits available for filling it = 2  {1,2}            (Since the first one cannot be 0)
Number of ways of filling the remaining four palaces = 3 each  (as each place can be filled with either 1, 2 or 0)
By fundamental principle of counting, number of five digit numbers that can be formed = 2×3×3×3×3 = 2×34

Page No 16.45:

Question 3:

In how many ways 4 women draw water from 4 taps, if no tap remains unused?

Answer:

Number of ways to draw water from the 1st tap = Number of women available to draw water = 4
Number of ways to draw water from the 2nd tap = Number of women available to draw water = 3
Number of ways to draw water from the 3rd tap = Number of women available to draw water = 2
Number of ways to draw water from the 4th tap = Number of women available to draw water = 1
∴ Total number of ways = 4×3×2×1 = 4! = 24

Page No 16.45:

Question 4:

Write the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number.

Answer:

Total number of outcomes when 3 dice are thrown = 6×6×6 = 216
Number of outcomes in which there is an odd number on all the three dice = 3×3×3 = 27
∴ Number of outcomes in which there is an even number at least on one dice = {Total possible outcomes} - {Number of outcomes in which there is an odd number on all the three dice } = 216 - 27 = 189

Page No 16.45:

Question 5:

Write the number of arrangements of the letters of the word BANANA in which two N's come together.

Answer:

The word BANANA consists of 6 letters including three As and two Ns.
Considering both Ns together or as a single letter, we are left with 5 letters including three As.
∴ Number of arrangements of 5 things in which 3 are similar to one kind = 5!3! = 20

Page No 16.45:

Question 6:

Write the number of ways in which 7 men and 7 women can sit on a round table such that no two women sit together.

Answer:


Each of the seven men can be arranged amongst themselves in 7! ways.
The women can be arranged amongst themselves in seven places, in 6! ways (i.e. n things can be arranged in (n-1)! ways around a round table).
By fundamental principle of counting, total number of ways = 7!×6!

Page No 16.45:

Question 7:

Write the number of words that can be formed out of the letters of the word 'COMMITTEE'.

Answer:


The word COMMITTEE consists of 9 letters including two Ms, two Ts and two Es.
Number of words that can be formed out of the letters of the word COMMITTEE
= Number of arrangements of 9 things of which 2 are similar to the first kind,
2 are similar to the second kind and 2 are similar to the third kind = 9!2!2!2!=9!2!3

Page No 16.45:

Question 8:

Write the number of all possible words that can be formed using the letters of the word 'MATHEMATICS'.

Answer:

The word 'MATHEMATICS' consists of 11 letters including two Ms, two Ts and two As
Number of words that can be formed out of the letters of the word MATHEMATICS = Number of arrangements of 11 things of which 2 are similar to the first kind, 2 are similar to the second kind and 2 are similar to the third kind = 11!2!2!2!

Page No 16.45:

Question 9:

Write the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together.

Answer:

Each of the six men can be arranged amongst themselves in 6! ways.
The five women can be arranged amongst themselves in the six places in 5! ways.
∴ By fundamental principle of counting, total number of ways = 6!×5!

Page No 16.45:

Question 10:

Write the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.

Answer:


Five boys can be arranged amongst themselves in 5! ways, at the places shown above.
The three girls are now to be arranged in the remaining four places taken three at a time = 4P3 = 4!
By fundamental principle of counting, total number of ways = 5!×4! = 120×24 = 2880

Page No 16.45:

Question 11:

Write the remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14.

Answer:

Every number after 6! (i.e. 7! onwards) till 200! will consist a power of 2 and 7, which will be exactly divisible by 14.

So, we need to divide only the sum till 6!.

1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873

When 873 is divided, the remainder would be same as when 1! + 2! + 3! + ... + 200! is divided by 14.
Remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14 = Remainder obtained when 873 is divided by 14 = 5

Page No 16.45:

Question 12:

Write the number of numbers that can be formed using all for digits 1, 2, 3, 4.

Answer:

Disclaimer:- (1) Here, we can form 4 digits, 5 digits , 6 digits numbers and so on.... using the given digits. Thus, infinite numbers can be formed.
(2) Taking into account only four digit numbers.

We have to find all the numbers that can be formed by using the digits 1, 2, 3 and 4. This means that repetition of digits is not allowed as all the digits have to be used.
Total numbers that can be formed = Number of arrangements of four digits, taken all at a time =  4! = 24



Page No 16.46:

Question 1:

The number of permutations of n different things taking r at a time when 3 particular things are to be included is
(a) n3Pr − 3

(b) n3Pr

(c) nPr − 3

(d) r ! n3Cr − 3

Answer:

(d) r ! n3Cr − 3
Here, we have to permute n things of which 3 things are to be included.
So, only the remaining (n-3) things are left for permutation, taking (r-3) things at a time. This is because 3 things have already been included.
But, these r things can be arranged in r! ways.
∴ Total number of permutations = r ! n3Cr − 3

Page No 16.46:

Question 2:

The number of five-digit telephone numbers having at least one of their digits repeated is
(a) 90000
(b) 100000
(c) 30240
(d) 69760

Answer:

(d) 69760
Total number of five digit numbers (since there is no restriction of the number 0XXXX) = 10×10×10×10×10 = 100000
These numbers also include the numbers where the digits are not being repeated. So, we need to subtract all such numbers.
Number of 5 digit numbers that can be formed without any repetition of digits = 10×9×8×7×6 = 30240
∴ Number of five-digit telephone numbers having at least one of their digits repeated = {Total number of 5 digit numbers} - {Number of numbers that do not have any digit repeated} = 100000 - 30240 = 69760

Page No 16.46:

Question 3:

The number of words that can be formed out of the letters of the word "ARTICLE" so that vowels occupy even places is
(a) 574
(b) 36
(c) 754
(d) 144

Answer:

(d) 144
The word ARTICLE consists of 3 vowels that have to be arranged in the three even places. This can be done in 3! ways.
And, the remaining 4 consonants can be arranged among themselves in 4! ways.
∴ Total number of ways = 3!×4! = 144

Page No 16.46:

Question 4:

How many numbers greater than 10 lacs be formed from 2, 3, 0, 3, 4, 2, 3 ?
(a) 420
(b) 360
(c) 400
(d) 300

Answer:

(b) 360
10 lakhs consists of seven digits.
 Number of arrangements of seven numbers of which 2 are similar of first kind, 3 are similar of second kind = 7!2!3!
But, these numbers also include the numbers in which the first digit has been considered as 0. This will result in a number less than 10 lakhs. Thus, we need to subtract all those numbers.
Numbers in which the first digit is fixed as 0 = Number of arrangements of the remaining 6 digits = 6!2!3!
Total numbers greater than 10 lakhs that can be formed using the given digits = 7!2!3!-6!2!3!
                                                                                                              = 420-60
                                                                                                              = 360

Page No 16.46:

Question 5:

The number of different signals which can be given from 6 flags of different colours taking one or more at a time, is
(a) 1958
(b) 1956
(c) 16
(d) 64

Answer:

(b) 1956
Number of permutations of six signals taking 1 at a time = 6P1
Number of permutations of six signals taking 2 at a time = 6P2
Number of permutations of six signals taking 3 at a time = 6P3
Number of permutations of six signals taking 4 at a time = 6P4
Number of permutations of six signals taking 5 at a time = 6P5
Number of permutations of six signals taking all at a time = 6P6

∴ Total number of signals  =6!5!+6!4!+6!3!+6!2!+6!1!+6! 
 
                                             = 6 + 30 + 120 + 360 + 720 + 720 = 1956

Page No 16.46:

Question 6:

The number of words from the letters of the word 'BHARAT' in which B and H will never come together, is
(a) 360
(b) 240
(c) 120
(d) none of these.

Answer:

(b) 240
Total number of words that can be formed of the letters of the word BHARAT = 6!2!
                                                                                                                = 360
Number of words in which the letters B and H are always together  = 2× 5!2!
                                                                                                 = 120
∴ Number of words in which the letters B and H are never together = 360-120
                                                                                                 = 240

Page No 16.46:

Question 7:

The number of six letter words that can be formed using the letters of the word "ASSIST" in which S's alternate with other letters is
(a) 12
(b) 24
(c) 18
(d) none of these.

Answer:

(a) 12
All S's can be placed either at even places or at odd places, i.e. in 2 ways.
The remaining letters can be placed at the remaining places in 3!, i.e. in 6 ways.
∴ Total number of ways = 6×2 = 12

Page No 16.46:

Question 8:

The number of arrangements of the word "DELHI" in which E precedes I is
(a) 30
(b) 60
(c) 120
(d) 59

Answer:

(b) 60
There are 4 cases where E precedes I i.e.
Case 1: When E and I are together, which are possible in 4 ways whereas other 3 letters are arranged in 3!,
So, the number of arrangements=4×3!=24
Case 2: When E and I have 1 letter in between, which are possible in 3 ways whereas other 3 letters are arranged in 3!,
So,the number of arrangements=3×3!=18
Case 3: When E and I have 2 letters in between, which are possible in 2 ways whereas other 3 letters are arranged in 3!,
So,the number of arrangements=2×3!=12
Case 4: When E and I have 3 letters in between, which are possible in 1 way whereas other 3 letters are arranged in 3!,
So,the number of arrangements=1×3!=6
Thus, total number of arrangements=24+18+12+6=60

Page No 16.46:

Question 9:

The number of ways in which the letters of the word 'CONSTANT' can be arranged without changing the relative positions of the vowels and consonants is
(a) 360
(b) 256
(c) 444
(d) none of these.

Answer:

(a) 360
The word CONSTANT consists of two vowels that are placed at the 2nd and 6th position, and six consonants.
The two vowels can  be arranged at their respective places, i.e. 2nd and 6th place, in 2! ways.
The remaining 6 consonants can be arranged at their respective places in 6!2!2!ways.
∴ Total number of arrangements = 2!×6!2!2! = 360

Page No 16.46:

Question 10:

The number of ways to arrange the letters of the word CHEESE are
(a) 120
(b) 240
(c) 720
(d) 6

Answer:

(a) 120
Total number of arrangements of the letters of the word CHEESE = Number of arrangements of 6 things taken all at a time, of which 3 are of one kind = 6!3! = 120

Page No 16.46:

Question 11:

Number of all four digit numbers having different digits formed of the digits 1, 2, 3, 4 and 5 and divisible by 4 is
(a) 24
(b) 30
(c) 125
(d) 100

Answer:

(a) 24
In order to make a number divisible by 4, its last two digits must be divisible by 4, which in this case can be 12, 24, 32 or 52.
Since repetition of digits is not allowed, the remaining first two digits can be arranged in 3×2 ways in each case.
∴ Total number of numbers that can be formed = 4×{3×2} = 24

Page No 16.46:

Question 12:

If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
(a) 324
(b) 341
(c) 359
(d) none of these

Answer:

(a) 324
When arranged alphabetically, the letters of the word KRISNA are A, I, K, N, R and S.

Number of words that will be formed with A as the first letter = Number of arrangements of the remaining 5 letters
                                                                                         = 5!
Number of words that will be formed with I as the first letter = Number of arrangements of the remaining 5 letters
                                                                                        = 5!
∴ The number of words beginning with KA = Number of arrangements of the remaining 4 letters
                                                               = 4!
The number of words starting with KI = Number of arrangements of the remaining 4 letters
                                                       = 4!
The number of words starting with KN = Number of arrangements of the remaining 4 letters
                                                        = 4!

Alphabetically, the next letter will be KR.
Number of words starting with KR followed by A, i.e. KRA = Number of arrangements of the remaining 3 letters
                                                                                      = 3!
Number of words starting with KRI followed by A, i.e. KRIA = Number of arrangements of the remaining 2 letters
                                                                                        = 2!
Number of words starting with KRI followed by N, i.e. KRIN = Number of arrangements of the remaining 2 letters
                                                                                         = 2!
The first word beginning with KRIS is the word KRISAN and the next word is KRISNA.

∴ Rank of the word KRISNA = 5! + 5! + 4! + 4! + 4! + 3! + 2! + 2! + 2
                                           = 324

Page No 16.46:

Question 13:

If in a group of n distinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects, then the number of objects is
(a) 10
(b) 8
(c) 6
(d) none of these.

Answer:

(c) 6
According to the question:
nP4 = 12×nP2
n!n-4!=12×n!n-2!n-2!n-4!=12n-2n-3  = 4×3n-2 =4n=6



Page No 16.47:

Question 14:

The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is
(a) 4! × 3!
(b) 4!
(c) 3! × 3!
(d) none of these.

Answer:

(a) 4! × 3!
According to the question, 3 men have to be 'consecutive' means that they have to be considered as a single man.
But, these 3 men can be arranged among themselves in 3! ways.
And, the remaining 3 men, along with this group, can be arranged among themselves in 4! ways.
∴ Total number of arrangements =  4! × 3!

Page No 16.47:

Question 15:

A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is
(a) 216
(b) 600
(c) 240
(d) 3125

Answer:

(a) 216
A number is divisible by 3 when the sum of the digits of the number is divisible by 3.
Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3.
1+2+3+4+5 = 15
0+1+2+4+5 = 12
Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers that can be formed = 5!
Similarly, using the digits 0, 1, 2, 4 and 5, the number that can be formed = 5!-4! {since the first digit cannot be 0}
∴ Total numbers that are possible = 5! + 5! - 4! = 240 - 24 = 216

Page No 16.47:

Question 16:

The product of r consecutive positive integers is divisible by
(a) r !
(b) (r − 1) !
(c) (r + 1) !
(d) none of these.

Answer:

(a) r !
The product of r consecutive integers  is equal to r!, so it will be divisible by r!.

Page No 16.47:

Question 17:

If k + 5Pk + 1 = 11 (k-1)2. k + 3Pk , then the values of k are
(a) 7 and 11
(b) 6 and 7
(c) 2 and 11
(d) 2 and 6

Answer:

(b) 6 and 7

 k + 5Pk + 1 = 11 (k-1)2. k + 3Pk

k+5!k+5-k-1!=11k-12×k+3!k+3-k!k+5!4!=11k-12×k+3!3!k+5!k+3!=11k-12×4!3!k+5k+4 = 22k-1k2+9k+20 = 22k-22k2-13k+42 = 0k = 6,7

Page No 16.47:

Question 18:

The number of arrangements of the letters of the word BHARAT taking 3 at a time is
(a) 72
(b) 120
(c) 14
(d) none of these.

Answer:

(a) 72

When we make words after selecting letters of the word BHARAT, it could consist of a single A, two As or no A.

Case-I: A is not selected for the three letter word.

Number of arrangements of three letters out of B, H, R and T = 4×3×2 = 24

Case-II: One A is selected and the other two letters are selected out of B, H, R or T.
Possible ways of selection: Selecting two letters out of B, H, R or T can be done in P24 =12 ways.
Now, in each of  these 12 ways, these two letters can be placed at any of the three places in the three letter word in 3 ways.

∴ Total number of words that can be formed = 12×3 = 36

Case-III: Two A's and a letter from B, H, R or T are selected.

Possible ways of arrangement: 
Number of ways of selecting a letter from B, H, R or T = 4

And now this letter can be placed in any one of the three places in the three letter word other than the two A's in 3 ways.

∴ Total number of words having 2 A's = 4×3 = 12

Hence, total number of words that can be formed = 24 + 36 + 12 = 72 

Page No 16.47:

Question 19:

The number of words that can be made by re-arranging the letters of the word APURBA so that vowels and consonants are alternate is
(a) 18
(b) 35
(c) 36
(d) none of these.

Answer:

(c) 36
The word APURBA is a 6 letter word consisting of 3 vowels that can be arranged in 3 alternate places, in 3!2!ways.
The remaining 3 consonants can be arranged in the remaining 3 places in 3! ways.
∴ Total number of words that can be formed = 3!2!×3! = 18
But this whole arrangement can be set-up in total two ways, i.e either  VCVCVC or CVCVCV.
∴ Total number of words = 18×2 = 36

Page No 16.47:

Question 20:

The number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons, is
(a) 60 × 5!
(b) 15 × 4! × 5!
(c) 4! × 5!
(d) none of these.

Answer:

(a) 60 × 5! 
The four people, i.e A, B and the two persons between them are always together. Thus, they can be considered as a single person.
So, along with the remaining 4 persons, there are now total 5 people who need to be arranged. This can be done in 5! ways.
But, the two persons that have to be included between A and B could be selected out of the remaining 6 people in 6P2 ways, which is equal to 30.
For each selection, these two persons standing between A and B can be arranged among themselves in 2 ways.
∴ Total number of arrangements = 5!×30×2 = 60×5!

Page No 16.47:

Question 21:

The number of ways in which the letters of the word ARTICLE can be arranged so that even places are always occupied by consonants is
(a) 576
(b) 4C3 × 4!
(c) 2 × 4!
(d) none of these.

Answer:

(a) 576
There are 3 even places in the 7 letter word ARTICLE.
So, we have to arrange 4 consonants in these 3 places in 4P3 ways.
And the remaining 4 letters can be arranged among themselves in 4! ways.
∴ Total number of ways of arrangement  = 4P3 ×4! = 4!×4! = 576

Page No 16.47:

Question 22:

In a room there are 12 bulbs of the same wattage, each having a separate switch. The number of ways to light the room with different amounts of illumination is
(a) 122 − 1
(b) 212
(c) 212 − 1
(d) none of these.

Answer:

(c) 212 − 1
Each of the bulb has its own switch, i.e each bulb will have two outcomes − it will either glow or not glow.
Thus, each of the 12 bulbs will have 2 outcomes.
∴ Total number of ways to illuminate the room = 212

Here, we have also considered the way in which all the bulbs are switched-off. However, this is not required as we need to find out only the number of ways of illuminating the room.
Hence, we subtract that one way from the total number of ways.
= 212 − 1



Page No 16.5:

Question 9:

If (n + 3)! = 56 [(n + 1)!], find n.

Answer:

(n + 3)! = 56 [(n + 1)!]
(n + 3)×(n + 2)×(n + 1)! = 56 [(n + 1)!]
(n + 3)×(n + 2) = 56
(n + 3)×(n + 2) = 8×7
n + 3 = 8
n = 5

Page No 16.5:

Question 10:

If (2n)!3! (2n-3)! and n!2! (n-2)! are in the ratio 44 : 3, find n.

Answer:

(2n)!3!(2n-3)!:n!2!(n-2)!=44:3  (2n)!3!(2n-3)!×2!(n-2)!n!=443(2n)(2n-1)(2n-2) [(2n-3)!]3(2!)(2n-3)!×2!(n-2)!n(n-1) [(n-2)!]=443(2n)(2n-1)(2n-2)3×1n(n-1) = 443(2n)(2n-1)(2)(n-1)3×1n(n-1) = 4434(2n-1)n(n-1)3×1n(n-1) = 4434(2n-1)3= 443(2n-1) = 112n = 12n = 6

Page No 16.5:

Question 11:

Prove that:
(i) n!(n-r)! = n (n − 1) (n − 2) ... (n − (r − 1))

(ii) n!(n-r)! r!+n!(n-r+1)! (r-1)!=(n+1)!r! (n-r+1)!

Answer:

(i) LHS= n!(n-r)!             =nn-1n-2n-3n-4 ... n-r+1n-r!(n-r)!             =nn-1n-2n-3n-4 ... n-r+1             =nn-1n-2n-3n-4 ... n-r-1 = RHS

(ii)   LHS = n!n-r!r!+n!n-r+1!          =n!n-r!r!+ n!(n-r+1) [(n-r)!]          =n!n-r+1+ n!r!r!n-r+1 [(n-r)!]          =n!n+1-n!r! + n!r!r!n-r+1n-r!          =n!(n+1)r!n-r+1n-r!          =n+1!r!n-r+1! = RHS  Hence proved.

Page No 16.5:

Question 12:

Prove that: (2n+1)!n! = 2n [1 · 3 · 5 ... (2n − 1) (2n + 1)]

Answer:

  LHS = 2n +1!n!            =2n+12n2n-1....4321n!          =135.........2n-12n+1246.........2nn!           =2n135.........2n-12n+1123.........nn!          =2n135.........2n-12n+1n!n!          =2n135.........2n-12n+1 = RHS   Hence, proved.



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