Rd Sharma Xi 2018 Solutions for Class 11 Science Math Chapter 17 Combinations are provided here with simple step-by-step explanations. These solutions for Combinations are extremely popular among Class 11 Science students for Math Combinations 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 11 Science Math Chapter 17 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 Class 11 Science Math are prepared by experts and are 100% accurate.

Page No 17.15:

Question 1:

From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?

Answer:

Required number of ways = 15C11
 
Now,15C11 =15C4                                                    
            = 154×143×132×121×11C0                
  
             = 1365                                                  

Page No 17.15:

Question 2:

How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?

Answer:

Clearly, out of the 25 boys and 10 girls, 5 boys and 3 girls will be chosen.

Then, different boat parties of 8 = 25C5×10C3
                                   
                                    =25!5! 20!×10!3! 7!= 25×24×23×22×215×4×3×2×1×10×9×83×2×1= 6375600

Page No 17.15:

Question 3:

In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?

Answer:

We are given that 2 courses are compulsory out of the 9 available courses,
Thus, a student can choose 3 courses out of the remaining 7 courses.
Number of ways = 7C3 = 7!3! 4! = 7×6×5×4!3×2×1×4! = 35

Page No 17.15:

Question 4:

In how many ways can a football team of 11 players be selected from 16 players? How many of these will
(i) include 2 particular players?
(ii) exclude 2 particular players?

Answer:

Number of ways in which 11 players can be selected out of 16 = 16C11 = 16!11! 5! = 16×15×14×13×125×4×3×2×1 = 4368
(i) If 2 particular players are included, it would mean that out of 14 players, 9 players are selected.
 Required number of ways = 14C9 = 14!9! 5! = 14×13×12×11×105×4×3×2×1 = 2002

(ii) If 2 particular players are excluded, it would mean that out of 14 players, 11 players are selected.
 Required number of ways = 14C11= 14!11! 3! = 14×13×123×2×1 = 364

Page No 17.15:

Question 5:

There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees:
(i) a particular professor is included.
(ii) a particular student is included.
(iii) a particular student is excluded.

Answer:

Clearly, 2 professors and 3 students are selected out of 10 professors and 20 students, respectively.
    Required number of ways  = 10C2×20C3 = 102×91×203×192×181 = 51300

(i) If a particular professor is included, it means that 1 professor is selected out of the remaining 9 professors.
  Required number of ways = 20C3×9C1 = 203×192×181×91 = 10260

(ii) If a particular student is included, it means that 2 students are selected out of the remaining 19 students.
   Required number of ways = 19C2×10C2 = 192×181×102×91 = 7695

(iii) If a particular student is excluded, it means that 3 students are selected out of the remaining 19 students.
  Required number of ways = 19C3×10C2 = 193×182×171×102×91 = 43605

Page No 17.15:

Question 6:

How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?

Answer:

Required number of ways of getting different products = 4C2+4C3 +4C4 = 6 + 4 + 1 = 11



Page No 17.16:

Question 7:

From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition; at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?

Answer:

Two girls who won the prizes last year are to be included in every selection.
So, we have to select 8 students out of 12 boys and 8 girls, choosing at least 4 boys and 2 girls.
Number of ways in which it can be done = 12C6×8C2 + 12C5×8C3 + 12C4×8C4 = 25872 + 44352 + 34650 = 104874

Page No 17.16:

Question 8:

How many different selections of 4 books can be made from 10 different books, if
(i) there is no restriction;
(ii) two particular books are always selected;
(iii) two particular books are never selected?

Answer:

(i) Required ways of selecting 4 books from 10 books without any restriction = 10C4 = 104×93×82×7 = 210

(ii) Two particular books are selected from 10 books. So, 2 books need to be selected from 8 books.
Required number of ways if 2 particular books are always selected = 8C2 = 82×71 = 28

(iii) Two particulars books are never selected from 10 books. So, 4 books need to be selected from 8 books.
Required number of ways if two particular books are never selected = 8C4  = 84×73×62×51 = 70

Page No 17.16:

Question 9:

From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?

Answer:

(i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer.
   Required number of ways = 4C1 × 8C5  = 4 × 8!5! 3! = 4×8×7×6×5!5! ×6 = 224

(ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer.
   Required number of ways =  Total number of ways - Number of ways in which no officer is selected
                                            =12C6  -8C6  = 12!6! 6! - 8!6! 2! = 12×11×10×9×8×76×5×4×3×2×1 - 8×72 = 924-28=896

Page No 17.16:

Question 10:

A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?

Answer:

A sports team of 11 students is to be constituted, choosing at least 5 students of class XI and at least 5 from class XII.
Required number of ways = 20C5×20C6 + 20C6×20C5 = 2 ×20C5×20C6 = 2 20C6 × 20C5

Page No 17.16:

Question 11:

A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions?

Answer:

The various possibilities for answering the 10 questions are given below:
(i) 4 from part A and 6 from part B.
(ii) 5 from part A and 5 from part B.
(iii) 6 from part A and 4 from part B.
∴ Required number of ways = 6C4×7C6 + 6C5×7C5 + 6C6×7C4
  
                                              =6!4! 2! × 7+ 6 × 7!5! 2! + 1 × 7!4! 3! =105+126+35=266

Page No 17.16:

Question 12:

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

Answer:

A student has to answer 4 question out of 5 questions.
Since questions 1 and 2 are compulsory, he/she will have to answer 2 question from the remaining 3.
∴ Required number of ways =3C2 = 3

Page No 17.16:

Question 13:

A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?

Answer:

Required ways = 6C5×6C2 +6C4×6C3 + 6C3×6C4 + 6C2×6C5
                       =26C5×6C2+6C4×6C3 =290+300=2390=780

Page No 17.16:

Question 14:

There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.

Answer:

Number of straight lines formed joining the 10 points, taking 2 points at a time = 10C2 = 102×91 = 45
Number of straight lines formed joining the 4 points, taking 2 points at a time = 4C2 = 42× 31 = 6
But, when 4 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines = 45- 6 +1 = 40

Page No 17.16:

Question 15:

Find the number of diagonals of (i) a hexagon (ii) a polygon of 16 sides.

Answer:

A polygon of n sides has n vertices. By joining any two vertices we obtain either a side or a diagonal.
∴ Number of ways of selecting 2 out of 9 =nC2=nn-12
Out of these lines, n lines are the sides of the polygon.

∴ Number of diagonals = nn-12-n=nn-32

(i) In a hexagon, there are 6 sides.
∴ Number of diagonals = 66-32=9 

(ii) There are 16 sides.
∴ Number of diagonals = 1616-32=104 

Page No 17.16:

Question 16:

How many triangles can be obtained by joining 12 points, five of which are collinear?

Answer:

Out of 12 points, 5 points are collinear and 3 points are required to form a triangle.
Required ways =12C3 -5C3  = 123×112×101 - 53×42×31 = 220 - 10 = 210

Page No 17.16:

Question 17:

In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?

Answer:

5 persons are to be selected out of 6 men and 4 women. At least, one woman has to be selected in all cases.

Required number of ways =4C1×6C4 + 4C2×6C3 + 4C3×6C2 +4C4×6C1= 60 + 120 + 60 + 6 = 246

Page No 17.16:

Question 18:

In a village, there are 87 families of which 52 families have at most 2 children. In a rural development programme, 20 families are to be helped chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made?

Answer:

52 families have at most 2 children, while 35 families have more than 2 children.
The selection of 20 families of which at least 18 families must have 2 children can be made in the ways given below.
(i) 18 families out of 52 and 2 families out of 35
(ii) 19 families out of 52 and 1 family out of 35
(iii) 20 families out of 52
∴ Required ways = 52C18 × 35C2  + 52C19 × 35C1  + 52C20 × 35C0 

Page No 17.16:

Question 19:

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls?

Answer:

A group consists of 4 girls and 7 boys. Out of them, 5 are to be selected to form a team.
(i) If the team has no girls, then the number of ways of selecting 5 members =7C5 = 7!5! 2! = 7×62 = 21

(ii) If the team has at least 1 boy and 1 girl, then the number of ways of selecting 5 members
= 4C1×7C4+4C2×7C3 + 4C3×7C2 +4C4×7C1 = 140 + 210+ 84 + 7 = 441    

(iii) If the team has at least 3 girls, then the number of ways of selecting 5 members
=4C3×7C2 +4C4×7C1 = 84 + 7 = 91

Page No 17.16:

Question 20:

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Answer:

A committee of 3 people is to be constituted from a group of 2 men and 3 women.
∴ Number of ways =2C0×3C3+2C1×3C2 + 2C2×3C1=1+ 2 × 3 + 3×1=10

Number of committees consisting of 1 man and 2 women =2C1×3C2 = 6

Page No 17.16:

Question 21:

Find the number of (i) diagonals (ii) triangles formed in a decagon.

Answer:

A decagon has 10 sides.
(i)  Number of diagonals = n n-32 = 10 10-32 = 35
(ii)  Number of triangles (i.e. 3 sides are to be selected)   = 10C3 = 103×92×81 = 120

Page No 17.16:

Question 22:

Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?

Answer:

There are 4 kings in the deck of cards.
So, we are left with 48 cards out of 52.
∴ Required combination =  48C1×4C4 + 48C2×4C3 + 48C3×4C2 + 48C4×4C1
                                       = 48 + 4512 + 103776 + 778320= 886656

Page No 17.16:

Question 23:

We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can the selection be made?

Answer:

6 people are to be selected from 8.

There are two case.
 (i) When A is selected, then B must be chosen.
Number of ways= 6C4=15

(ii) When A is not chosen:
Number of ways=7C6=7

∴ Total number of ways = 15 + 7 = 22

Page No 17.16:

Question 24:

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer:

3 boys and 3 girls are to be selected from 5 boys and 4 girls.
∴ Required ways =  5C3×4C3 = 53×42×31×43×32×21 =40

Page No 17.16:

Question 25:

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer:

Required number of ways = 6C3×5C3×5C3 = 6!3! 3!× 5!3! 2!×5!3! 2! = 2000

Page No 17.16:

Question 26:

Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Answer:

There are total 4 aces in the deck of 52 cards. So, we are left with 48 cards.
∴ Required ways = 4C1×48C4= 41×484×473×462×451 = 778320



Page No 17.17:

Question 27:

In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer:

Out of 17 players, 11 need to be selected. There are 5 bowlers, of which four must be selected in the team. So, we have to choose 7 players from the remaining 12 players.
Required number of ways = 5C4×12C7 = 5× 12!7! 5! = 5×792 = 3960 

Page No 17.17:

Question 28:

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Answer:

2 black and 3 red balls are to be selected from 5 black and 6 red balls.
Required number of ways = 5C2×6C3 = 52×41×63×52×41 = 200

Page No 17.17:

Question 29:

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

2 courses are compulsory out of the 9 available courses. There are 7 more courses.
So, we need to choose 3 courses out of 7 courses.
∴ Required number of ways = 7C3 = 73×62×51 = 35

Page No 17.17:

Question 30:

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) at most 3 girls?

Answer:

 A committee of 7 has to be formed from 9 boys and 4 girls.

(i) When the committee consists of exactly 3 girls:
    Required number of ways =  4C3×9C4 = 43×32×21×94×83×72×61 = 504

(ii) When the committee consists of at least 3 girls:
    Required number of ways = 4C3×9C4 + 4C4×9C3 =504 + 84 = 588

(iii) When the committee consists of at most 3 girls:

 Required number of ways = 4C0×9C7 + 4C1×9C6 + 4C2×9C5 +4C3×9C4 = 36 + 336 + 756 + 504 = 1632

Page No 17.17:

Question 31:

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Answer:

A question paper consists of 12 questions divided into 2 parts, one with 5 and the other with 7 questions.
A student has to attempt 8 questions out of the 12 questions by selecting at least 3 from each part.
∴ Required number of ways =5C3×7C5 + 5C4×7C4+5C5×7C3 = 210 + 175 + 35 = 420

Page No 17.17:

Question 32:

A parallelogram is cut by two sets of m lines parallel to its sides. Find the number of parallelograms thus formed.

Answer:

Each set of parallel lines consists of m+2 lines.
Each parallelogram is formed by choosing two lines from the first set and two straight lines from the second set.
∴ Total number of parallelograms = m+2C2  × m+2C2   = m+2C22

Page No 17.17:

Question 33:

Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many (i) straight lines (ii) triangles can be formed by joining them?

Answer:

(i) Number of straight lines formed joining the 18 points, taking 2 points at a time = 18C2 = 182×171 = 153
Number of straight lines formed joining the 5 points, taking 2 points at a time = 5C2 = 52× 41 = 10
But, when 5 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines = 153 - 10 +1 = 144

(ii) Number of triangles formed joining the 18 points, taking 3 points at a time = 18C3 = 183×172×161 = 816
Number of straight lines formed joining the 5 points, taking 3 points at a time = 5C3 = 53× 42×31 = 10
∴ Required number of triangles = 816 - 10= 806



Page No 17.23:

Question 1:

How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?

Answer:

2 out of 5 vowels and 3 out of 17 consonants can be chosen in5C2× 17C3 ways.
Thus, there are5C2× 17C3 groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters, which can be arranged in 5! ways.
∴ Required number of words = 5C2× 17C35! = 6800 ×120 = 816000

Page No 17.23:

Question 2:

There are 10 persons named P1, P2, P3, ...., P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.

Answer:

We need to arrange 5 persons in a line out of 10 persons, such that in each arrangement P1 must occur whereas P4 and P5 do not occur.

First we choose 5 persons out of 10 persons, such that in each arrangement P1 must occur whereas P4 and P5 do not occur.

Number of such selections = 7C4

Now, in each selection 5 persons can be arranged among themselves in 5! ways.

∴ required number of arrangements = 7C4 × 5! = 7×6×53×2×1×5×4×3×2×1=4200

Thus, ​number of such possible arrangements is 4200.

Page No 17.23:

Question 3:

How many words, with or without meaning can be formed from the letters of 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 letter is a vowel?

Answer:

There are six letters in the word MONDAY.

(i) 4 letters are used at a time:
Four letters can be chosen out of six letters in 6C4 ways.
So, there are 6C4 groups containing four letters that can be arranged in 4! ways.
∴ Number of ways =6C4× 4! = 6!4! 2!× 4! = 6!2! = 360

(ii) All the letters are used at a time:
This can be done in 6C6 ways.
So, there are 6C6 groups containing six letters that can be arranged in 6! ways.
∴ Number of ways =6C6× 6! = 1× 720 = 720

(iii) All the letters are used, but the first letter is a vowel:
There are two vowels, namely A and O, in the word MONDAY.
For the first letter, out of the two vowels, one vowel can be chosen in 2C1 ways.
The remaining five letters can be chosen in 5C5 ways.
So, the letters in 5C5 group can be arranged in 5! ways.
∴ Number of ways =2C1×5C5× 5! = 2×1×5! = 240

Page No 17.23:

Question 4:

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Answer:

Given r places, we first fill up 3 places by 3 particular things. This can be done in rPways.

Now, we have to fill remaining r − 3 places with remaining − 3 things.

This can be done in n − 3Pr − 3 ways.

Thus, the required number of permutations will be rP3 × n − 3Pr − 3 ways.

Page No 17.23:

Question 5:

How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Answer:

There are 4 vowels and 4 consonants in the word INVOLUTE.
Out of these, 3 vowels and 2 consonants can be chosen in 4C3×4C2 ways.
The 5 letters that have been selected can be arranged in 5! ways.
∴ Required number of words = 4C3× 4C2 × 5! = 4×6×120 = 2880

Page No 17.23:

Question 6:

Find the number of permutations of n different things taken r at a time such that two specified things occur together?

Answer:

We have n different things.
We are to select r things at a time such that two specified things occur together.
Remaining things = n -2
Out of the remaining (n -2) things, we can select (r -2) things in n -2Cr -2 ways.
Consider the two things as one and mix them with (r -2) things.
Now, we have (r -1) things that can be arranged in (r -1)! ways.
But, two things can be put together in 2! ways.

 Required number of ways = n-2Cr-2×r-1!×2!                                              =2r-1n-2Cr-2×r-2!                                              =2r-1n-2Pr-2

Page No 17.23:

Question 7:

Find the number of ways in which : (a) a selection (b) an arrangement, of four letters can be made from the letters of the word 'PROPORTION'.

Answer:

There are 10 letters in the word PROPORTION, namely OOO, PP, RR, I, T and N.
(a) The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the second kind
(iii) 2 alike letters and 2 distinct letters
(iv) all distinct letters

Now, we shall discuss these four cases one by one.
(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, OOO, which can be selected in one way.
Out of the 5 different letters, P, R, I, T and N, one can be selected in5C1 = 5 ways.
(ii) There are 3 sets of two alike letters, which can be selected in 3C2 = 3 ways.
(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 3C1×5C2 = 30 ways.
(iv) There are 6 different letters.
Number of ways of selecting 4 letters = 6C4 = 15

∴ Total number of ways = 5+ 3 + 30 + 15 = 53

(b) The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the second kind
(iii) 2 alike letters and 2 distinct letters
(iv) all distinct letters

Now, we shall discuss these four cases one by one.
(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, OOO, which can be selected in one way.
Out of the 5 different letters, P, R, I, T and N, one can be selected in5C1 ways.
These four letters can be arranged in 4!3! 1! ways.
∴ Total number of ways =5C1×4!3! 1!=20
(ii) There are 3 sets of two alike letters, which can be selected in 3C2 ways.
Now, the letters of each group can be arranged in 4!2! 2! ways.
∴ Total number of ways =3C2×4!2! 2!=18
(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 3C1×5C2 = 30 ways.
Now, the letters of each group can be arranged in 4!2! ways.
∴ Total number of ways = 30×4!2!=360
(iv) There are 6 different letters.
So, the number of ways of selecting 4 letters is 6C4 = 15 and these letters can be arranged in 4! ways.
∴ Total number of ways = 15 × 4! = 360

∴ Total number of ways = 20 + 18 + 360 + 360 = 758

Page No 17.23:

Question 8:

How many words can be formed by taking 4 letters at a time from the letters of the word 'MORADABAD'?

Answer:

There are 9 letters in the word MORADABAD, namely AAA, DD, M, R, B and O.
The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the other kind
(iii) 2 alike letters and 2 distinct letters
(iv) all different letters

(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, AAA, which can be selected in one way.
Out of the 5 different letters D, M, R, B and O, one can be selected in5C1 ways.
These four letters can be arranged in 4!3! 1! ways.
∴ Total number of ways =5C1×4!3! 1!=20

(ii) There are two sets of two alike letters, which can be selected in 2C2 ways.
Now, the letters of each group can be arranged in 4!2! 2! ways.
∴ Total number of ways =2C2×4!2! 2!=6

(iii) There is only one set of two alike letters, which can be selected in 2C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 2C1×5C2 = 20 ways.
Now, the letters of each group can be arranged in 4!2! ways.
∴ Total number of ways = 20×4!2!=240

(iv) There are 6 different letters A, D, M,B, O and R.
So, the number of ways of selecting 4 letters is 6C4, i.e. 15, and these letters can be arranged in 4! ways.
∴ Total number of ways = 15 × 4! = 360
∴ Total number of ways = 20 + 6 + 240 + 360 = 626

Page No 17.23:

Question 9:

A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?

Answer:

A businessman hosts a dinner for 21 guests.
15 people can be accommodated at one table in 21C15 ways. They can arrange themselves in 15-1!=14! ways.
The remaining 6 people can be accommodated at another table in 6C6 ways. They can arrange themselves in 6-1!=5! ways.
∴ Total number of ways =21C15×6C6×14!×5!=21C15×14!×5!

Page No 17.23:

Question 10:

Find the number of combinations and permutations of 4 letters taken from the word 'EXAMINATION'.

Answer:

There are 11 letters in the word EXAMINATION, namely AA, NN, II, E, X, M, T and O.
The four-letter word may consist of
(i) 2 alike letters of one kind and 2 alike letters of the second kind
(ii) 2 alike letters and 2 distinct letters
(iii) all different letters
Now, we shall discuss the three cases one by one.

(i) 2 alike letters of one kind and 2 alike letters of the second kind:
There are three sets of 2 alike letters, namely AA, NN and II.
Out of these three sets, two can be selected in 3C2 ways.
So, there are 3C2 groups, each containing 4 letters out of which two are alike letters of one kind and two 2 are alike letters of the second kind.
Now, 4 letters in each group can be arranged in 4!2! 2! ways.
∴ Total number of words that consists of 2 alike letters of one kind and 2 alike letters of the second kind =3C2×4!2! 2! = 3×6 = 18

(ii) 2 alike and 2 different letters:
Out of three sets of two alike letters, one set can be chosen in 3C1 ways.
Now, from the remaining 7 letters, 2 letters can be chosen in 7C2 ways.
Thus, 2 alike letters and 2 distinct letters can be chosen in 3C1×7C2 ways.
So, there are 3C1×7C2  groups of 4 letters each.
Now, the letters in each group can be arranged in 4!2! ways. 
∴ Total number of words consisting of 2 alike and 2 distinct letters = 3C1× 7C2 ×4!2! = 756
(iii) All different letters:
There are 8 different letters, namely A, N, I, E, X, M, T and O. Out of them, 4 can be selected in 8C4 ways.
So, there are 8C4 groups of 4 letters each. The letters in each group can be arranged in 4! ways.
∴ Total number of four-letter words in which all the letters are distinct =8C4× 4! = 1680

∴ Total number of four-letter words = 18 + 756 + 1680 = 2454

Page No 17.23:

Question 11:

A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on the other side. In how many ways can they be seated?

Answer:

A tea party is arranged for 16 people along two sides of a long table with 8 chairs on each side.
4 people wish to sit on side A (say) and two on side B (say).
Now, 10 people are left, out of which 4 people can be selected for side A in 10C4 ways.
And, from the remaining people, 6 people can be selected for side B in 6C6 ways.
∴ Number of selections  =10C4× 6C6
Now, 8 people on each side can be arranged in 8! ways.
∴ Total number ways in which the people can be seated  =10C4× 6C6× 8! × 8! = 10C4×8!2



Page No 17.24:

Question 1:

Write r=0m Crn+r in the simplified form.

Answer:

We know:
Crn+Cr-1n=Crn+1

r=0mCrn+r=C0n+C1n+1+C2n+2+C3n+3+...+Cmn+m C0n=C0n+1 r=0mCrn+r=C0n+1+C1n+1+C2n+2+C3n+3+...+Cmn+mUsing Cr-1n+Crn=Crn+1:r=0mCrn+r=C1n+2+C2n+2+C3n+3+...+Cmn+mr=0mCrn+r=C2n+3+C3n+3+...+Cmn+m

Proceeding in the same way:

r=0mCrn+r=Cm-1n+m+Cmn+m=Cmn+m+1r=0mCrn+r=Cmn+m+1

Page No 17.24:

Question 2:

If 35Cn +7 = 35C4n − 2 , then write the values of n.

Answer:

 35Cn +7 = 35C4n − 2 
n+7+4n-2=35           [∵nCx=nCyn=x+y or x=y]                          
5n=30n=6And, n+7=4n-2       n=3

Page No 17.24:

Question 3:

Write the number of diagonals of an n-sided polygon.

Answer:

An n-sided  polygon has n vertices.
By joining any two vertices of the polygon, we obtain either a side or a diagonal of the polygon.
Number of line segments obtained by joining the vertices of an n-sided polygon if we take two vertices at a time = Number of ways of selecting 2 out of n = nC2
Out of these lines, n lines are sides of the polygon.
Number of diagonals of the polygon = nC2 - n = n (n-1)2 - n = n(n-3)2

Page No 17.24:

Question 4:

Write the expression nCr +1 + nCr − 1 + 2 × nCr in the simplest form.

Answer:

nCr+1+ nCr-1 +2.nCr 
   
 = nCr+1 +nCr + nCr +nCr-1      nCr +nCr-1=n+1Cr=n+1Cr+1 +n+1Cr             nCr +nCr-1=n+1Cr=n+2Cr+1       

Page No 17.24:

Question 5:

Write the value of r=16 C356-r + C450.

Answer:

We know:
nCr-1 + nCr = n+1Cr
Now, we have:                     r=16 56-rC3+ 50C4                 =55C3+54C3+53C3+52C3+51C3+50C3+50C4
                   
                 =55C3+54C3+53C3+52C3+51C3+51C4=55C3+54C3+53C3+52C3+52C4=55C3+54C3+53C3+53C4=55C3+54C3+54C4=55C3+55C4=56C4

Page No 17.24:

Question 6:

There are 3 letters and 3 directed envelopes. Write the number of ways in which no letter is put in the correct envelope.

Answer:

Total number of ways in which the letters can be put = 3! = 6
Suppose, out of the three letters, one has been put in the correct envelope.
This can be done in 3C1, i.e. 3, ways.
Now, out of three, if two letters have been put in the correct envelope, then the last one has been put in the correct envelope as well.
This can be done in 3C3, i.e. one way.
∴ Number of ways = 3 + 1 = 4
∴ Number of ways in which no letter is put in the correct envelope = 6-4 = 2

Page No 17.24:

Question 7:

Write the maximum number of points of intersection of 8 straight lines in a plane.

Answer:

We know that two lines are required for one point of intersection.
∴ Number of points of intersection =8C2=82×71=28

Page No 17.24:

Question 8:

Write the number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.

Answer:

A parallelogram is formed by choosing two straight lines from a set of four parallel lines and two straight lines from a set of three parallel lines.
Two straight lines from the set of four parallel lines can be chosen in 4C2 ways and two straight lines from the set of three parallel lines can be chosen in 3C2 ways.
∴ Number of parallelograms that can be formed = C24 × C23 = 4!2! 2! × 3!2! 1! = 6 × 3 = 18

Page No 17.24:

Question 9:

Write the number of ways in which 5 red and 4 white balls can be drawn from a bag containing 10 red and 8 white balls.

Answer:

4 white and 5 red balls are to be selected from 8 white and 10 red balls.
∴ Required number of ways =8C4×10C5

Page No 17.24:

Question 10:

Write the number of ways in which 12 boys may be divided into three groups of 4 boys each.

Answer:

Number of groups in which 12 boys are to be divided = 3
Now, 4 boys can be chosen out of 12 boys in C4×8C4×4C124 ways.
These groups can be arranged in 3! ways. 
∴ Total number of ways = 12C4× 8C4× 4C43! = 12!×8!4!×8!×4!×4!×3!=12!4!3×3!

Page No 17.24:

Question 11:

Write the total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants.

Answer:

2 out of 4 vowels and 3 out of 5 consonants can be chosen in4C2× 5C3 ways.
The total number of letters is 5. These letters can be arranged in 5! ways.
∴ Total number of words =4C2× 5C3× 5! 



Page No 17.25:

Question 1:

If 20Cr = 20Cr−10, then 18Cr is equal to
(a) 4896
(b) 816
(c) 1632
(d) nont of these

Answer:

(b) 816
 r + r - 10 = 20                          [∵nCx=nCyn=x+y or x=y ]            
2r - 10 = 202r = 30 r = 15

Now,18Cr=18C15
    18C15 = 18C3                                         
   18C3 = 183×172×16 = 816  

Page No 17.25:

Question 2:

If 20Cr = 20Cr + 4 , then rC3 is equal to
(a) 54
(b) 56
(c) 58
(d) none of these

Answer:

(b) 56
 r + r + 4 = 20                         [∵nCx=nCy n=x+y or x=y]
 2r + 4 = 20 2r = 16 r = 8

Now,rC3 = 8C3
8C3 = 8!3! 5! = 8×7×63×2×1 = 56

Page No 17.25:

Question 3:

If 15C3r = 15Cr + 3 , then r is equal to
(a) 5
(b) 4
(c) 3
(d) 2

Answer:

(c) 3
3r + r + 3 = 15                       [∵nCx=nCy n=x+y or x=y]
 4r + 3 = 15 4r = 12 r = 3

Page No 17.25:

Question 4:

If 20Cr + 1 = 20Cr − 1 , then r is equal to
(a) 10
(b) 11
(c) 19
(d) 12

Answer:

(a) 10
r + 1 + r - 1 = 20                           [∵nCx=nCy n=x+y or x=y]
 2r = 20 r = 10 

Page No 17.25:

Question 5:

If C (n, 12) = C (n, 8), then C (22, n) is equal to
(a) 231
(b) 210
(c) 252
(d) 303

Answer:

(a) 231
nC12 =nC8
 n = 12 + 8 = 20                                        [∵nCx=nCy n=x+y or x=y]
   
Now,22Cn =22C20
                                                          
               = 222×211                            
                = 231

Page No 17.25:

Question 6:

If mC1 = nC2 , then
(a) 2 m = n
(b) 2 m = n (n + 1)
(c) 2 m = n (n − 1)
(d) 2 n = m (m − 1)

Answer:

(c) 2 m = n (n − 1)

    mC1 = nC2
 m!1! m-1! = n!2! n-2! m m-1!m-1! = n n-1 n-2!2 n-2!  2m = n n-1
       

Page No 17.25:

Question 7:

If nC12 = nC8 , then n =
(a) 20
(b) 12
(c) 6
(d) 30

Answer:

(a) 20

n = 12 + 8 = 20                  [∵nCx=nCy n=x+y or x=y]

Page No 17.25:

Question 8:

If nCr + nCr + 1 = n + 1Cx , then x =
(a) r
(b) r − 1
(c) n
(d) r + 1

Answer:

(d) r + 1

nCr + nCr+1 = n+1Cx   [Given]

We have:
               nCr +nCr+1 = n+1Cx                           [∵nCr +nCr-1 = n+1Cr]
                 n+1Cr+1 = n+1Cx 
               r+1= x                                          [∵nCx=nCy n=x+y or x=y]

Page No 17.25:

Question 9:

If C2(a2-a)=C4(a2-a) , then a =
(a) 2
(b) 3
(c) 4
(d) none of these

Answer:

(b) 3

       a2-a =2 + 4                     [∵nCx=nCyn=x+y or x=y]
 a2 - a - 6 = 0a2 - 3a + 2a - 6 = 0 a a-3 + 2 a-3 = 0 a+2 a-3 = 0
 a=-2 or a=3
But, a=-2 is not possible.
a = 3

Page No 17.25:

Question 10:

5C1 + 5C2 + 5C3 + 5C4 +5C5 is equal to
(a) 30
(b) 31
(c) 32
(d) 33

Answer:

(b) 31

       5C1 + 5C2 + 5C3 + 5C4 + 5C5
 = 5C1 + 5C2 + 5C2 + 5C1 + 5C5                  

 = 2 × 5C1 + 2 × 5C2 + 5C5= 2 × 5 + 2 × 5!2! 3! + 1  = 10 + 20 + 1 = 31                         

Page No 17.25:

Question 11:

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
(a) 60
(b) 120
(c) 7200
(d) none of these

Answer:

 (c) 7200
2 out of 4 vowels can be chosen in 4C2 ways and 3 out of 5 consonants can be chosen in 5C3 ways.
Thus, there are C2× 5C43  groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters that can be arranged in 5! ways.
∴ Required number of words = 4C2× 5C3× 5! = 60 × 120 = 7200

Page No 17.25:

Question 12:

There are 12 points in a plane. The number of the straight lines joining any two of them when 3 of them are collinear, is
(a) 62
(b) 63
(c) 64
(d) 65

Answer:

(c) 64
Number of straight lines joining 12 points if we take 2 points at a time = 12C2 = 12!2! 10! = 66

Number of straight lines joining 3 points if we take 2 points at a time = 3C2 = 3

But, 3 collinear points, when joined in pairs, give only one line.
∴ Required number of straight lines = 66 -3 + 1 = 64

Page No 17.25:

Question 13:

Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?
(a) 60
(b) 20
(c) 15
(d) 125

Answer:

(a) 60
Three persons can take 5 seats in 5C3 ways. Moreover, 3 persons can sit in 3! ways.
∴ Required number of ways =  5C3× 3! = 10×6 = 60

Page No 17.25:

Question 14:

In how many ways can a committee of 5 be made out of 6 men and 4 women containing at least one women?
(a) 246
(b) 222
(c) 186
(d) none of these

Answer:

(a) 246

  Required number of ways=4C1× 6C4 +4C2× 6C3  +4C3× 6C2  +4C4× 6C1                                               = 60 + 120 + 60 + 6                                              = 246



Page No 17.26:

Question 15:

There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two of them is
(a) 45
(b) 40
(c) 39
(d) 38

Answer:

(b) 40
Number of straight lines formed by joining the 10 points if we take 2 points at a time =10C2 = 102×91 = 45
Number of straight lines formed by joining the 4 points if we take 2 points at a time =4C2 = 42× 31 = 6
But, 4 collinear points, when joined in pairs, give only one line.
∴ Required number of straight lines = 45 - 6 +1 = 40

Page No 17.26:

Question 16:

There are 13 players of cricket, out of which 4 are bowlers. In how many ways a team of eleven be selected from them so as to include at least two bowlers?
(a) 72
(b) 78
(c) 42
(d) none of these

Answer:

(b) 78

4 out of 13 players are bowlers.
In other words, 9 players are not bowlers.
A team of 11 is to be selected so as to include at least 2 bowlers.

 Number of ways =4C2×9C9 + 4C3×9C8 + 4C4×9C7                              = 6 + 36 + 36                              = 78

Page No 17.26:

Question 17:

If C0 + C1 + C2 + ... + Cn = 256, then 2nC2 is equal to
(a) 56
(b) 120
(c) 28
(d) 91

Answer:

(b) 150

If set S has n elements, then C n, k is the number of ways of choosing k elements from S.
Thus, the number of subsets of S of all possible values is given by
    Cn,0 + Cn,1 + Cn,3 + ... + Cn,n = 2n
Comparing the given equation with the above equation:
       2n = 256 2n = 28 n = 8

  2nC2 = 16C216C2  = 16!2! 14! = 16×152 = 120

Page No 17.26:

Question 18:

The number of ways in which a host lady can invite for a party of 8 out of 12 people of whom two do not want to attend the party together is
(a) 2 × 11C7 + 10C8
(b) 10C8 + 11C7
(c) 12C810C6
(d) none of these

Answer:

(c) 12C810C6

A host lady can invite 8 out of 12 people in 12C8 ways. Two out of these 12 people do not want to attend the party together.
∴ Number of ways = 12C810C6

Page No 17.26:

Question 19:

Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the number of circles that can be drawn so that each contains at least 3 of the given points is
(a) 216
(b) 156
(c) 172
(d) none of these

Answer:

(b) 156
We need at least three points to draw a circle that passes through them.
Now, number of circles formed out of 11 points by taking three points at a time = 11C3 = 165
Number of circles formed out of 5 points by taking three points at a time = 5C3 = 10
It is given that 5 points lie on one circle.

Required number of circles = 165- 10 + 1 = 156

Page No 17.26:

Question 20:

How many different committees of 5 can be formed from 6 men and 4 women on which exact 3 men and 2 women serve?
(a) 6
(b) 20
(c) 60
(d) 120

Answer:

(d) 120

Number of committes that can be formed = 6C3 × 4C2                                                                 = 6!3! 3! × 4!2! 2!                                                                 = 6 × 5 × 43 × 2 × 4 × 32                                                                 = 120

Page No 17.26:

Question 21:

If 43Cr − 6 = 43C3r + 1 , then the value of r is
(a) 12
(b) 8
(c) 6
(d) 10
(e) 14

Answer:

(a) 12

 r - 6 + 3r + 1 = 43                             [∵nCx=nCy n=x+y or x=y]
 4r - 5 = 43  4r = 48 r = 12

Page No 17.26:

Question 22:

The number of diagonals that can be drawn by joining the vertices of an octagon is
(a) 20
(b) 28
(c) 8
(d) 16

Answer:

(a) 20

An octagon has 8 vertices. 
The number of diagonals of a polygon is given by n n-32.
∴ Number of diagonals of an octagon = 8 8-32 = 20

Page No 17.26:

Question 23:

The value of C07+C17+C17+C27+...+C67+C77 is
(a) 27 − 1
(b) 28 − 2
(c) 28 − 1
(d) 28

Answer:

(b) 28 − 2


   C07+C17+C17+C27+C27+C37+C37+C47+C47+C57+C57+C67+C67+C77

=1 + 2 × C17 + 2 × C27 + 2 × C37 + 2 × C47 + 2 × C57 + 2 × C67 + 1    

=1 +2 × C17 +  2 × C27 + 2 × C37 + 2 × C37 + 2 × C27 + 2 × C67 + 1      

= 2 + 22 C17 + C27 + C37 = 2 + 22 7 + 72 × 6 + 73 × 62 × 5

= 2 + 252 = 254 = 28 - 2

Page No 17.26:

Question 24:

Among 14 players, 5 are bowlers. In how many ways a team of 11 may be formed with at least 4 bowlers?
(a) 265
(b) 263
(c) 264
(d) 275

Answer:

(c) 264

Among 14 players, 5 are bowlers.
A team of 11 players has to be selected such that at least 4 bowlers are included in the team.
 Required number of ways=C45×C79 + C55× C69                                              = 180 + 84                                              = 264

Page No 17.26:

Question 25:

A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is
(a) 112
(b) 140
(c) 164
(d) none of these

Answer:

(b) 140

Suppose there are two friends, A and B, who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting 6 guests = 8C6 = 28
If one of them attends the party, then the number of ways of selecting 6 guests = 2.8C5 = 112
∴ Total number of ways = 112 + 28 = 140

Page No 17.26:

Question 26:

If n + 1C3 = 2 · nC2 , then n =
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) 5
            C3n+1 =2×C2n n+1!3! n-2! = 2× n!2! n-2! n+1 n!3×2! n-2! = 2× n!2! n-2! n+1 = 6 n = 5

Page No 17.26:

Question 27:

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is
(a) 6
(b) 9
(c) 12
(d) 18

Answer:

(d) 18
A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines.
Two parallel lines from the set of four parallel lines can be chosen in 4C2 ways and two parallel lines from the set of 3 parallel lines can be chosen in 3C2 ways.
∴ Number of parallelograms that can be formed = C24× C23 = 4!2! 2!×3!2! 1! =6×3 = 18



Page No 17.8:

Question 1:

Evaluate the following:
(i) 14C3
(ii) 12C10
(iii) 35C35
(iv) n + 1Cn
(v) r=15 5Cr.

Answer:

(i) We have,                                                                                                       
 14C3= 143 × 132× 121×11C0        [∵ nCr = nr n-1Cr-1]
 14C3 = 364                             [∵ nC0 = 1]

(ii) We have,
         12C10=12C2                                                            [∵ nCr = nCn-r]
 
     12C10=12C2 = 122×111×10C0                            [∵ nCr = nr n-1Cr-1]

   12C10 = 122×111× 1                                               [∵ nC0 = 1]

   12C10 = 66

(iii) We have,
        
      35C35 = 35C0                                                     [∵ nCr = nCn-r]

       35C35 =1                                                          [∵ nC0 = 1]


(iv) We have,
     
       n + 1Cn =n + 1C1                                               [∵ nCr = nCn-r]
n+1Cn= n+1C1=  n+11× nC0                          [∵ nCr = nr n-1Cr-1]
   n + 1Cn = n+1                                                  [∵ nC0 = 1]

(v) We have,

      r=15 5Cr=5C1+5C2+5C3+5C4+5C5

    r=15 5Cr=5C1+5C3+5C3+5C1+5C0                   [∵ nCr = nCn-r]

    r=155Cr = 2×51×4C0 +2×53×42×31×2C0 + 5C0     [∵ nCr = nr n-1Cr-1]

  r=155Cr = 10 + 20 + 1 = 31.                                                [∵ nC0 = 1]
   

Page No 17.8:

Question 2:

If nC12 = nC5, find the value of n.

Answer:

We have,
nC12 = nC5
 n = 12 + 5 = 17                     [∵   nCx = nCy   x = y or, n = x+y]

Page No 17.8:

Question 3:

If nC4 = nC6, find 12Cn.

Answer:

We have,
  nC4= nC6   
 n = 6+4 = 10                                                [∵   nCx = nCy   x = y or, n = x+y]

   Now, 12C10= 12C2                                          [∵ nCr = nCn-r]
12C10 =12C2 = 122×111×10C0                     [∵ nCr = nr n-1Cr-1]
 12C10 = 66                                                     [∵ nC0 = 1]

 

Page No 17.8:

Question 4:

If nC10 = nC12, find 23Cn.

Answer:

Given: nC10 = nC12
We have,
nC10 = nC12          
 n = 12+10 = 22                                   [∵   nCx = nCy   x = y or, n = x+y]

Now, 23C22 = 23C1                                       [∵ nCr = nCn-r]
 23C22  =23C1  = 231×22C0                   [∵ nCr = nr n-1Cr-1]
23C22  = 23                                            [∵ nC0 = 1]

Page No 17.8:

Question 5:

f 24Cx = 24C2x + 3, find x.

Answer:

Given:
 24Cx = 24C2x + 3
We have,
     24 = x + 2x + 3                           [∵ nCx = nCy   x = y or, n = x+y]
 24 = 3x + 33x = 21x = 7

Page No 17.8:

Question 6:

If 18Cx = 18Cx + 2, find x.

Answer:

Given: 
18Cx = 18Cx + 2

By using  nCx = nCy   x = y or  n = x+y we get,
18 = x + x + 2 2x = 16 x = 8

Page No 17.8:

Question 7:

If 15C3r = 15Cr + 3, find r.

Answer:

Given:
 15C3r = 15Cr + 3
     15 = 3r + r + 3.                 [∵ nCx = nCy   x = y or, n = x+y]
 15 = 4r + 3 4r = 12 r = 3

Page No 17.8:

Question 8:

If 8Cr7C3 = 7C2, find r.

Answer:

Given:
 8Cr7C3 = 7C2
We have,
8Cr= 7C2 + 7C3
 8Cr=8C3                                  [∵  nCr + nCr-1 = n+1Cr ; rn]
r = 3                                         [∵ nCx = nCy   x = y or, n = x+y]

And r+3=8r=5

 

Page No 17.8:

Question 9:

If 15Cr : 15Cr − 1 = 11 : 5, find r.

Answer:

Given:
 15Cr : 15Cr − 1 = 11 : 5

We have,
     15Cr15Cr-1 = 115

 15-r+1r  = 115                 [∵ nCrnCr-1 = n-r+1r]

 75 - 5r + 5 = 11r 16r = 80 r = 5

Page No 17.8:

Question 10:

If n +2C8 : n2P4 = 57 : 16, find n.

Answer:

We have, n +2C8 : n2P4 = 57 : 16

 n+2C8n-2P4 = 5716 (n+2)! 8! (n-6)!×(n-6)!(n-2)! = 5716(n+2) (n+1) n (n-1) (n-2)!8!×1(n-2)! = 5716 (n+2) (n+1) n (n-1) = 5716×8! = 19×316×8×7×6×5×4×3×2×1(n+2) (n+1) n (n-1) = 143640 (n-1) n (n+1) (n+2) = 19×3×7×6×5×4×3(n-1) n (n+1) (n+2) = 19×(3×7)×(6×3)×(4×5)(n-1) n (n+1) (n+2) = 18×19×20×21 n-1 = 18 n = 19

Page No 17.8:

Question 11:

If 28C2r : 24C2r − 4 = 225 : 11, find r.

Answer:

We have,  28C2r : 24C2r − 4 = 225 : 11
 28C2r24C2r-4 = 22511 28!2r!  (28-2r)!× (2r-4)! (28-2r)!24! = 22511 28×27×26×252r (2r-1) (2r-2) (2r-3) = 22511 2r (2r-1) (2r-2) (2r-3) = 28×27×26×25×11225 2r (2r-1) (2r-2) (2r-3) = 28×3×26×11 2r (2r-1) (2r-2) (2r-3) = 4×7×3×13×2×11 2r (2r-1) (2r-2) (2r-3) = (2×7)×13×(3×4)×11 2r (2r-1) (2r-2) (2r-3) = 14×13×12×11 2r = 14r = 7

Page No 17.8:

Question 12:

If nC4 , nC5 and nC6 are in A.P., then find n.

Answer:

Since nC4 , nC5 and nC6 are in AP.

∴ 2. nC5 = nC4 + nC6
2×n!5!n-5!=n!4!n-4!+n!6!n-6!25×4!n-5n-6!=14!n-5n-4n-6!+16×5×4!n-6!25n-5=1n-5n-4+13025n-5-1n-5n-4=1302n-8-55n-5n-4=1302n-13n-5n-4=1612n-78=n2-9n+20n2-21n+98=0n-7n-14=0 n=7 and 14

Page No 17.8:

Question 13:

If 2nC3 : nC2 = 44 : 3, find n.

Answer:

Given: 2nC3:nC2 = 44:3
     2nC3nC2 = 4432n!3! (2n-3)!× 2! (n-2)!n! = 443 2n (2n-1) (2n-2)3 n (n-1) = 443 (2n-1) (2n-2) = 22 (n-1) 4n2 - 6n + 2 = 22n - 22 4n2 - 28n + 24 = 0 n2 - 7n + 6 = 0 n2 - 6n - n + 6 = 0 n (n-6) -1(n-6) = 0 (n-1) (n-6) = 0
 n =1   or,   n= 6

Now,  n=1  2C3:2C2 = 44:3 
But, this is not possible.
∴  n = 6

Page No 17.8:

Question 14:

If 16Cr = 16Cr + 2, find rC4.

Answer:

Given:
 16Cr = 16Cr+2

    16 = r + r + 2                               [∵ Property 5: nCx = nCy  x = y   or   x + y = n]
2r + 2 = 16 2r = 14 r = 7

Now,rC4 = 7C4
 7C4 = 7C3                                            [∵ nCr =nCn-r]
7C4 = 7C3 = 73×62×51×4C0            [∵ nCr = nr . n-1Cr-1 ]
7C4= 35                                             [∵ nC0 = 1]

Page No 17.8:

Question 15:

If α = mC2, then find the value of αC2.

Answer:

αC2 = α2×(α-1)1×αC0                           [∵ nCr = nr. n-1Cr-1]
         
          = 12 α (α-1)                                      [∵ nC0 = 1]

         = 12 mC2 mC2 -1=12 m!2! m-2! m!2! m-2! -1= 12 m m-12 m m-12 -1=12 mm-12 m m-1 - 22 = 18 m m-1 m m-1 -2=18 m2 m-12 - 2m m -1= 18 m2 m2 + 1 - 2m -2m2 + 2m= 18 m4 + m2 - 2m3 - 2m2 + 2m= 18 m4 -2m3 - m2 + 2m= 18 m2 - 2m m2 -1= 18 m m-2 m-1 m+1= 18 m+1 m m-1 m-2

          
         

Page No 17.8:

Question 16:

Prove that the product of 2n consecutive negative integers is divisible by (2n)!

Answer:

Let 2n negative integers be -r,-r-1, -r-2,....,..., -r-2n+1.

Then, product = -12nrr+1r+2,....,...r+2n-1
              = r-1! rr+1 r+2 ......r+2n-1r-1!= r+2n-1!r-1!= r+2n-1!r-1!2n!×2n!= r+2n-1C2n×2n!
This is divisible by 2n!.           

Page No 17.8:

Question 17:

For all positive integers n, show that 2nCn + 2nCn − 1 = 12(2n + 2Cn + 1).

Answer:

LHS =  2nCn+ 2nCn-1       =2n!n! n! + 2n!n-1! 2n - n +1!       = 2n!n! n! + 2n!n-1! n+1!       = 2n!n n-1! n! + 2n!n-1! n+1n!       = 2n!n! n-1! 1n + 1n+1       = 2n!n! n-1! 2n+1n n+1       = 2n+1!n! n+1!

RHS = 12 2n+2Cn+1        = 12 2n+2!n+1! 2n + 2 - n-1!        = 12 2n+2!n+1! n+1!        = 12 2n+2 2n+1!n+1 n! n+1!        = 12 2n+1 2n+1!n+1 n! n+1!        = 2n+1!n! n+1!

∴ LHS = RHS

Page No 17.8:

Question 18:

Prove that: 4nC2n : 2nCn = [1 · 3 · 5 ... (4n − 1)] : [1 · 3 · 5 ... (2n − 1)]2.

Answer:

4nC2n2nCn=1.3.5...4n-11.3.5...2n-12LHS=4nC2n2nCn        =4n!2n!2n!×n!n!2n!       =4n×4n-1×4n-2×4n-3.................3×2×1 ×n!22n×2n-1×2n-2.......3×2××122n!      =1×3×5........4n-12×4×6...............4n×n!21×3×5×.........2n-122×4×6×.......2n2×2n!    =1×3×5........4n-1×22n×1×2×3..........2nn!21×3×5×.........2n-12×22n1×2×3×.......n22n!   =1×3×5........4n-12n!n!21×3×5×.........2n-12n!22n!   =1×3×5........4n-11×3×5×.........2n-12 =RHSHence, proved.

Page No 17.8:

Question 19:

Evaluate C520+r=25 25-rC4.

Answer:

Given:
 20C5 + r=25 25-rC4

 20C5 + r=25 25-rC4= 20C5+23C4 + 22C4 + 21C4 + 20C4=20C4 + 20C5 +21C4+22C4+23C4

 =21C5 +21C4  + 22C4+23C4                                 [∵ nCr-1 +nCr =n+1Cr]
 =21C4 + 21C5 +22C4+23C4
 =22C5+22C4+23C4                                               [∵nCr-1 +nCr =n+1Cr]
 =23C5+22C4+23C4
 =23C5+23C4=24C5                                                     [∵ nCr-1 +nCr =n+1Cr]
                                                                 
 =25!19! 5! = 24×23×22×21×205×4×3×2×1 = 42504



Page No 17.9:

Question 20:

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a) CrnCr-1n=n-r+1r
(b) n · n1Cr − 1 = (nr + 1) nCr − 1
(c) CrnCr-1n-1=nr
(iv) nCr + 2 · nCr − 1 + nCr − 2 = n + 2Cr.

Answer:

(a)
 nCrnCr-1 = n-r+1r
 
LHS = nCrnCr-1         = n!r! n-r!×r-1! n-r+1!n!         = n-r+1 n-r! r-1!r r-1! n-r!        = n-r+1r = RHS

∴  LHS = RHS

(b)
LHS= n. n-1Cr-1        = n n-1!r-1! n-1-r+1!         = n!r-1! n-r!RHS=n-r+1 nCr        = n-r+1 n!r-1! n-r+1!         =n-r+1n!r-1! n-r+1n-r!         = n!r-1! n-r!

∴ LHS = RHS

(c)
 nCrn-1Cr-1 = nr

 LHS = nCrn-1Cr-1         = n!r! n-r!×r-1! n-1-r+1!n-1!         =n n-1!r r-1! n-r! × r-1! n-r!n-1!         = nr = RHS

∴  LHS = RHS

(d)
nCr+ 2. nCr-1 +nCr-2 = n+2Cr

 LHS =nCr +2.nCr-1 +nCr-2       = nCr +nCr-1 + nCr-1 +nCr-2      
       = n+1Cr + n+1Cr-1                              [∵ nCr + nCr-1 = n+1Cr]
       =n+2Cr                                                [∵ nCr + nCr-1 = n+1Cr]     
       = RHS
∴  LHS = RHS



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