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#### 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?

Required number of ways = ${}^{15}{C}_{11}$

Now,

= 1365

#### 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?

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

Then, different boat parties of 8 = ${}^{25}{C}_{5}{×}^{10}{C}_{3}$

#### Question 3:

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

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 =

#### 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?

Number of ways in which 11 players can be selected out of 16 =
(i) If 2 particular players are included, it would mean that out of 14 players, 9 players are selected.
Required number of ways =

(ii) If 2 particular players are excluded, it would mean that out of 14 players, 11 players are selected.
Required number of ways =

#### 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.

Clearly, 2 professors and 3 students are selected out of 10 professors and 20 students, respectively.
Required number of ways  =

(i) If a particular professor is included, it means that 1 professor is selected out of the remaining 9 professors.
Required number of ways =

(ii) If a particular student is included, it means that 2 students are selected out of the remaining 19 students.
Required number of ways =

(iii) If a particular student is excluded, it means that 3 students are selected out of the remaining 19 students.
Required number of ways =

#### Question 6:

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

Required number of ways of getting different products =

#### 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?

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 =

#### 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?

(i) Required ways of selecting 4 books from 10 books without any restriction =

(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 =

(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 =

#### 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?

(i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer.
Required number of ways =

(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

#### 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?

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 =

#### 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?

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 =

#### 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.

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 =

#### 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?

Required ways =

#### 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.

Number of straight lines formed joining the 10 points, taking 2 points at a time =
Number of straight lines formed joining the 4 points, taking 2 points at a time =
But, when 4 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines =

#### Question 15:

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

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 ${=}^{n}{C}_{2}=\frac{n\left(n-1\right)}{2}$
Out of these lines, n lines are the sides of the polygon.

∴ Number of diagonals = $\frac{n\left(n-1\right)}{2}-n=\frac{n\left(n-3\right)}{2}$

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

(ii) There are 16 sides.
∴ Number of diagonals =

#### Question 16:

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

Out of 12 points, 5 points are collinear and 3 points are required to form a triangle.

#### 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?

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

#### 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?

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 =

#### 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?

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 =

(ii) If the team has at least 1 boy and 1 girl, then the number of ways of selecting 5 members

(iii) If the team has at least 3 girls, then the number of ways of selecting 5 members
=

#### 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?

A committee of 3 people is to be constituted from a group of 2 men and 3 women.
∴ Number of ways =

Number of committees consisting of 1 man and 2 women =

#### Question 21:

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

A decagon has 10 sides.
(i)  Number of diagonals =
(ii)  Number of triangles (i.e. 3 sides are to be selected)   =

#### 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?

There are 4 kings in the deck of cards.
So, we are left with 48 cards out of 52.
∴ Required combination =

#### 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?

6 people are to be selected from 8.

There are two case.
(i) When A is selected, then B must be chosen.

(ii) When A is not chosen:

∴ Total number of ways = 15 + 7 = 22

#### Question 24:

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

3 boys and 3 girls are to be selected from 5 boys and 4 girls.
∴ Required ways =

#### 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.

Required number of ways =

#### 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.

There are total 4 aces in the deck of 52 cards. So, we are left with 48 cards.
∴ Required ways =

#### 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?

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 =

#### 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.

2 black and 3 red balls are to be selected from 5 black and 6 red balls.
Required number of ways =

#### 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?

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 =

#### 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?

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 =

(ii) When the committee consists of at least 3 girls:
Required number of ways =

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

Required number of ways =

#### 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?

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 =

#### Question 32:

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

Each set of parallel lines consists of $\left(m+2\right)$ 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 =

#### 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?

(i) Number of straight lines formed joining the 18 points, taking 2 points at a time =
Number of straight lines formed joining the 5 points, taking 2 points at a time =
But, when 5 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines =

(ii) Number of triangles formed joining the 18 points, taking 3 points at a time =
Number of straight lines formed joining the 5 points, taking 3 points at a time =
∴ Required number of triangles =

#### Question 1:

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

2 out of 5 vowels and 3 out of 17 consonants can be chosen in ways.
Thus, there are groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters, which can be arranged in $5!$ ways.
∴ Required number of words =  =

#### Question 2:

There are 10 persons named . 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.

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! = $\frac{7×6×5}{3×2×1}×5×4×3×2×1=4200$

Thus, ​number of such possible arrangements is 4200.

#### 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?

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 =

(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 =

(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 =

#### Question 4:

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

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.

#### Question 5:

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

There are 4 vowels and 4 consonants in the word INVOLUTE.
Out of these, 3 vowels and 2 consonants can be chosen in $\left({}^{4}{C}_{3}{×}^{4}{C}_{2}\right)$ ways.
The 5 letters that have been selected can be arranged in 5! ways.
∴ Required number of words =

#### Question 6:

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

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.

#### 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'.

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 in${}^{5}{C}_{1}$ = 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 in${}^{5}{C}_{1}$ ways.
These four letters can be arranged in ways.
∴ Total number of ways =
(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 ways.
∴ Total number of 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.
Now, the letters of each group can be arranged in $\frac{4!}{2!}$ ways.
∴ Total number of ways = $30×\frac{4!}{2!}=360$
(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

#### Question 8:

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

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 in${}^{5}{C}_{1}$ ways.
These four letters can be arranged in ways.
∴ Total number of ways =

(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 ways.
∴ Total number of ways =

(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 $\frac{4!}{2!}$ ways.
∴ Total number of ways = $20×\frac{4!}{2!}=240$

(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

#### 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?

A businessman hosts a dinner for 21 guests.
15 people can be accommodated at one table in 21C15 ways. They can arrange themselves in $\left(15-1\right)!=14!$ ways.
The remaining 6 people can be accommodated at another table in 6C6 ways. They can arrange themselves in $\left(6-1\right)!=5!$ ways.
∴ Total number of ways =${}^{21}{C}_{15}{×}^{6}{C}_{6}×14!×5!{=}^{21}{C}_{15}×14!×5!$

#### Question 10:

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

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  ways.
∴ Total number of words that consists of 2 alike letters of one kind and 2 alike letters of the second kind =

(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 $\left({}^{3}{C}_{1}{×}^{7}{C}_{2}\right)$ ways.
So, there are $\left({}^{3}{C}_{1}{×}^{7}{C}_{2}\right)$  groups of 4 letters each.
Now, the letters in each group can be arranged in $\frac{4!}{2!}$ ways.
∴ Total number of words consisting of 2 alike and 2 distinct letters =
(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 =

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

#### 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?

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  =
Now, 8 people on each side can be arranged in $8!$ ways.
∴ Total number ways in which the people can be seated  =

#### Question 1:

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

(b) 816
[∵ ]

Now,${}^{18}{C}_{r}{=}^{18}{C}_{15}$

#### Question 2:

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

(b) 56
[∵]

Now,

#### Question 3:

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

(c) 3
[∵]

#### Question 4:

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

(a) 10
[∵]

#### Question 5:

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

(a) 231

[∵]

Now,

#### 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)

(c) 2 m = n (n − 1)

mC1 = nC2

#### Question 7:

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

(a) 20

[∵]

#### Question 8:

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

(d) r + 1

[Given]

We have:
[∵]

[∵]

#### Question 9:

If ${}^{\left({a}^{2}-a\right)}C_{2}={}^{\left({a}^{2}-a\right)}C_{4}$ , then a =
(a) 2
(b) 3
(c) 4
(d) none of these

(b) 3

[∵]

But, $a=-2$ is not possible.

#### Question 10:

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

(b) 31

#### 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

(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   groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters that can be arranged in 5! ways.
∴ Required number of words =

#### 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

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

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 =

#### 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

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

#### 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

(a) 246

#### 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

(b) 40
Number of straight lines formed by joining the 10 points if we take 2 points at a time =
Number of straight lines formed by joining the 4 points if we take 2 points at a time =
But, 4 collinear points, when joined in pairs, give only one line.
∴ Required number of straight lines =

#### 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

(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.

#### Question 17:

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

(b) 150

If set $S$ has n elements, then  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

Comparing the given equation with the above equation:

#### 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

(c) 12C810C6

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

#### 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

(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.

$\therefore$ Required number of circles = 165$-$ 10 + 1 = 156

#### 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

(d) 120

#### Question 21:

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

(a) 12

[∵]

#### 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

(a) 20

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

#### Question 23:

The value of $\left({}^{7}C_{0}+{}^{7}C_{1}\right)+\left({}^{7}C_{1}+{}^{7}C_{2}\right)+...+\left({}^{7}C_{6}+{}^{7}C_{7}\right)$ is
(a) 27 − 1
(b) 28 − 2
(c) 28 − 1
(d) 28

(b) 28 − 2

$\left({}^{7}C_{0}+{}^{7}C_{1}\right)+\left({}^{7}C_{1}+{}^{7}C_{2}\right)+\left({}^{7}C_{2}+{}^{7}C_{3}\right)+\left({}^{7}C_{3}+{}^{7}C_{4}\right)+\left({}^{7}C_{4}+{}^{7}C_{5}\right)+\left({}^{7}C_{5}+{}^{7}C_{6}\right)+\left({}^{7}C_{6}+{}^{7}C_{7}\right)$

#### 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

(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.

#### 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

(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

#### Question 26:

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

(c) 5

#### 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

(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 =

#### Question 28:

The number of ways in which a committee consisting of 3 men and 2 women, can be chose from 7 men and 5 women, is
(a) 45
(b) 350
(c) 4200
(d) 230

The selection of 5 members, consisting of 3 men and 2 women from 7 men and 5 women, can be made by selecting 3 men from 7 men and 2 women from 5 available
∴ This can be done in 7C3 × 5C

Hence, the correct answer is option B.

#### Question 29:

The number of signals that can be sent by 6 flags of different colours taking one or more at a time is
(a) 63
(b) 1956
(c) 720
(d) 21

Number of signals possible with one flag is 6.
Number of signals possible with 2 flags is 6P2 .
Number of signals possible with 3 flags is 6P3 .
Similarly with 4 flags 6P4, 5 flags is 6Pand with all 6 flags 6P.
∴ Total number of signals = 6P1 + 6P2 6P+  6P+ 6P+ 6P
= 6 + 30 + 120 + 360 + 720 + 720 = 1956

#### Question 30:

The straight lines l1, l2 and l3 are parallel and lie in the same plane. A total number of m points are taken on l1, n points on l2, k points on l3. The maximum number of triangles formed with vertices at these points are
(a) m + n + kC3
(b) m + n + kC3mC3nC3kC3
(c) mC3 + nC3 + kC3
(d) mC3 × nC3 × kC3

Here the total number of points are m + n + k .
Which must give m + n + kCnumbers of triangles.
Since m points lie l1, taking 3 points at a time is mC3.
similarly, from n points on l2, taking 3 points at a time is nC3 and kC3 from k
from these, three points mC3 or nC3 or kC3 triangle can not be formed
∴ The required number of triangles is m + n + kC− mC3  −  nC3  − kC3
Hence, the correct answer is option B.

#### Question 31:

The number of committees of five persons with a chairperson that can be formed from 12 persons, is
(a) 12C5
(b) 12C4
(c) 12 × 11C4
(d) 11C4

For committee consisting of 5 persons with are chair person, from 12 persons.
Chair persons can be selected in 12 ways.
Remaining 4 can be selected from remaining 11
Hence, number of committee of five person with a chair person is 12 × 11C4
Hence, the correct answer is option C.

#### Question 32:

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) 720

Number of vowels given is 4
Number of consonants given is 5
We have to form words by 2 vowels and 3 consonants number of  ways of selecting 2 vowels out of 4 is 4C2 number of  ways of selecting 3 consonants out of 5 is 5C3.
Hence number of ways of selection is 4C× 5C3

Now, there selected 5 letters can be arranged is 5! ways
So, total number of words is 60 × 5!
= 60 × 120
= 7200
Hence the correct answer is option C.

#### Question 33:

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

Five digit number is to be formed from 0,1,2,3,4 and 5.
Such that number is divisible by 3.
Any number is divisible by 3 if sum of its digits out of these digits,
1, 2, 3, 4, 5 and 0, 1, 2, 4, 5 sum up to be a multiple of 3
→ for 1,2,3,4,5,
The number of ways a five digit number which is divisible by 3 is 5 × 4 × 3 × 2 × 1                (∵ no restriction is there)
→ for 0,1, 2, 4, 5
The number of ways a five digit number which is divisible by 3 is 4 × 4 × 3 × 2 × 1 = 96     (∵ 0 cannot be placed or first place)
∴ Total number formed
= 120 + 96
= 216
Hence, the correct answer is option A.

#### Question 34:

Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
(a) 11
(b) 12
(c) 13
(d) 14

Let us suppose there are n persons in a room for a handshake, 2 people are requried who will shake hand with each other.
So, number of ways to make a pair out of n persons is nC2 = 66 (given)

Since n = −11 is not possible
⇒  n = 12
Hence, the correct answer is option B.

#### Question 35:

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
(a) 105
(b) 15
(c) 175
(d) 185

Number of ways of selecting 3 points from given 12 points = 12C3 .
But any three points selected from given seven collinear points does not from triangle number ways of selecting three points team seven collinear points = 7C3
∴ Required number of triangle = 12C−  7C3

Hence, the correct answer is option D.

#### Question 36:

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(a) 3600
(b) 3720
(c) 3800
(d) 3600

Number of ways of selecting one green dye is 5C1 + 5C2 + 5C3 + 5C4 + 5C5  = 25−1 ways.
Number of ways of selecting one blue dye can be choosen is 4C1 + 4C2 + 4C+ 4C4 = 24−1 ways.
and
Number of ways of selecting red dye can be choosen in 3C0 + 3C1 + 3C+ 3C3 = 23 ways.
So, total number of required selection
= (25 − 1) × (24−1) × 23
= 3720
Hence, the correct answer is option B.

#### Question 37:

The total number of 9 digit numbers which have all different digits is
(a) 10!
(b) 9!
(c) 9 × 9!
(d) 10 × 10!

Out of 10 digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9  _ _ _ _ _ _ _ _
First place has 9 options since 0 can not be place for number to be 9 digit.
Second place has 9 options since out of 10 are is already choosen.
Similarly third place has 8 options and so on...
∴ Number of 9 digit number with different digit is

i.e 9 × 9!

Hence, the correct answer is option C.

#### Question 38:

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) 18
(c) 12
(d) 9

Since parallelogram needs two set of parallel lines
i.e selecting two parallel lines from a set of four can be done is 4C2 =
and selecting two parallel lines from set of three parallel lines can be done in 3C2$\frac{3!}{2!}=3$
∴ Number of parallelogram that can be formed is 6 × 3 = 18.

#### Question 39:

The number of 5-digit telephone numbers having at least one of their digits repeated is
(a) 90,000
(b) 10,000
(c) 30240
(d) 69760

Total number of five digit telephone number possible without any restriction is 105 and number of 5 digit telephone number having all digits different is 10P5
∴ Required number of ways i.e having at-least one digit repeated is =105 − 10P5
i.e 1000000 − 10 × 9 × 8 × 7 × 6
= 105 − 30240
= 69760
Hence, the correct answer is option D.

#### Question 40:

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two man and exactly twice as many women as men, is
(a) 94
(b) 126
(C) 128
(d) none of these

Number of men = 4
Number of women = 6    (given)
Given condition says, committee include at-least two man and exactly twice as many women as men .
So, we can select either 2 men and 4 women or 3 men and 6 women.
∴ Required number of ways of forming committee is  4C2 × 6C4 + 4C3 × 6C6

Hence, the correct answer is option A.

#### Question 41:

The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them, is
(a) 16C11
(b) 16C5
(c) 16C6
(d) 20C9

Total number of players = 22
We need to select 11 players
Given condition says,
Exclude 4 particular player gives us 18 players available and including 2  particular players.
Implies only 9 need to be selected out of remaining 16.
Hence, number of ways of selection is 16C9.
Hence, the correct answer is option C.

#### Question 1:

If nPr = 840 and nCr = 35, then r = ___________.

Given  nPr = 840 and nCr = 35

#### Question 2:

The value of 15C815C915C615C7 is ___________.

15C8 + 15C915C15C7
Since nCr = nCnr
15C6 = 15C15−6 = 15C9     ...(1)
and 15C7 = 15C15−7 = 15C  ...(2)
15C8 + 15C9 − 15C9 − 15C8 = 0     (∵ from (1) and (2))

#### Question 3:

The value of nPr + nCr is ___________.

Therefore,  nPr ÷ nC = r

#### Question 4:

If n is even, then nCr is maximum when ____________.

If n is even,
for any n
The possible combinations are
nC0, nC1 , nC2 .........nCr, .........nCn−2, nCn−1, nC
i.e total n + 1 combinations are possible the value of coefficient increase initially, then reaches its maximum and start decreasing.
nCr is r + 1 th term and middle term is $\frac{n}{2}+1$
$\begin{array}{rcl}& & ⇒r+1=\frac{n}{2}+1\\ & & ⇒r=\frac{n}{2}\end{array}$
i.e r at which nCr is maximum.

#### Question 5:

If 2 × nC5 = 9 × n – 2C5, then n = ____________.

#### Question 6:

If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then r = ____________.

nCr-1 = 36, nCr = 84 and nCr + 1 = 126

#### Question 7:

If 18C15 + 2 (18C16) + 17C16 + 1 = nC3, then n = ____________.

Given18C15 + 2 (18C16) + 17C16 + 1 = nC3
L.H.S 18C15 + 18C16 18C16 + 17C16 + 1
Since, nCr+1 + nCr = n+1Cr+ and 1 = 17C17
L.H.S reduces to,
19C16 + 18C16  + 17C16 17C17
=19C16 + 18C16  + 18C17
= 19C16 + 19C17
i.e L.H.S = 20C17 and R.H.S = nC3 = L.H.S (given)
∴ 20C17 = nC
or 20C= nC   [∵ 20C17 = 20C20−17  = 20C3]
n = 20

#### Question 8:

If nC12 = nC6, then nC2 = ____________.

Given nC12 = nC6
Using nCr = nCn
n − 6 = 12

n = 18

i.e 18C = 153

#### Question 9:

If 189C35 + 189Cr = 190Cr, then r = ____________.

Given :- 189C35 + 189Cr = 190C
Using nCr + nCr+1 = n+1Cr+1
n = 89 and r = 36

#### Question 10:

If nP4 = 24. nC5, then the value of n is ____________.

Given :- nP4 = 24. nC5
Since

i.e n − 4 = 5
i.e n = 9

#### Question 11:

The value of nCr + 2 nCr – 1 + nCr – 2, 2 ≤ rn, is ____________.

nCr + 2nCr−1 + nCr−2
nCr + nCr−1 + nCr−1 + nCr−2
= n+1Cr + n+1Cr−1
= n+2C using identity
nCr + 2 nCr−1 + nCr−2 = n+2C2

#### Question 12:

A box contain 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ___________.

At-least 1 black ball can be selected in following ways
1 black ball and two non - black or 2 black ball and one non - black or all 3 black balls
∴ Total number of ways of selecting is

i.e The number of ways three balls be drawn from the box with at-lest one black ball included is 64.

#### Question 13:

Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if at least 2 are red is ___________.

Number of red balls = 5
Number of white balls = 4
Number of black balls = 3
Number of ball drawn = 3
Note, at-least 2 red balls can be drawn in following ways
→ 2 red and 1 non red.
→ all 3 reds balls.
∴ Number of ways of drawing at-least two red balls is all red 5C3 + 5C2 × 7C1                                                                                  $=\frac{4×5}{2}+\frac{4×5}{2}×7\phantom{\rule{0ex}{0ex}}=10+35×2\phantom{\rule{0ex}{0ex}}=80$

#### Question 14:

The total number of ways in which six '+' and four '–' signs can be arranged in aline such that no two '–' signs occur together is ___________.

Number of  '+' sign = 6 and Number of  '−' sign = 4
After placing 6 '+' sign, there are 7 places for '−' sign.
Now, we need to select 4 sign out of 7 places
∴ The total number of ways of arranging signs
Such that no two '−' are together = 7C4

#### Question 15:

A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. The number of different ways this can be done, if two particular women refuse to serve on the same committee is ___________.

Here, Number of men = 10
Number of women = 7
6 committee numbers can be selected containing at-least 3 men and 2 women in following 2 ways.
→ 4 men and 2 women
→ 3 men and 3 women
∴ number of ways of selecting at-least 3 men and 2 women in committee of 6

Number of ways when two particular women are never together = 10C3

Therefore, Total number of ways when two particular women are never together = 8610 − 810 = 7800.

#### Question 16:

The number of committees of five persons with a chair person can be selected from 12 persons, is ___________.

A chair person can be selected in 12 ways
Now, we need to select 4 person out of remaining 11.
∴ Number of committees of five person with a chair person can be selected in 12 × 11Cways

#### Question 17:

The number of automobile license plates that can be made if each plate contains two different letters of English alphabet followed by three distinct digits, is ___________.

Each automobile license plates has 5 places to fill 2 places of alphabet (out of 26) can be filled in 26 × 25 ways
Then 3 places of digits can be filled in 10 × 4 × 8 ways
Hence number of different automobile license
plates = 26 × 25 × 10 × 9 × 8

= 468000

#### Question 18:

The number of permutations of n distinct objects taken r at a time in which three particular objects occurs together is ___________.

Total number of objects (distinct) = n
Since, 3 objects are taken together (i.e 3 objects are together)
The number of ways of selecting r distinct objects from n is nCr
If three objects occur together, then number of ways = n−3Cr−3
also number of arrangements of these three things = 3!
and number of arrangements of (− 3 + 1) objects = (r − 2)!
∴ Total possible ways of taking r at a time from n objects such that three particular are together is n−3Cr−3 (r − 2)! 3!

#### Question 19:

Out of 10 persons P1, P2,..., P10, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. The number of such arrangements is ___________.

Since out of 5 to be selected, P1 is fix i.e P1 must occur.
⇒ we need to select 4 person out of 9 remaining.
Now, out of remaining 9, P4 and P5 do not occur
⇒ Available options are 7
⇒ Number of such arrangement
= 7C4 ×  5!   (since 5 persons in line can be arranged in 5! ways.)

#### Question 1:

Write in the simplified form.

We know:
${}^{n}C_{r}+{}^{n}C_{r-1}={}^{n+1}C_{r}$

Proceeding in the same way:

$\sum _{r=0}^{m}{}^{n+r}C_{r}={}^{n+m}C_{m-1}+{}^{n+m}C_{m}={}^{n+m+1}C_{m}\phantom{\rule{0ex}{0ex}}⇒\sum _{r=0}^{m}{}^{n+r}C_{r}={}^{n+m+1}C_{m}$

#### Question 2:

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

35Cn +7 = 35C4n − 2
$n+7+4n-2=35$           [∵]

#### Question 3:

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

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 =

#### Question 4:

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

#### Question 5:

Write the value of .

We know:
nCr$-$1 + nCr = n+1Cr

${=}^{55}{C}_{3}{+}^{54}{C}_{3}{+}^{53}{C}_{3}{+}^{52}{C}_{3}{+}^{51}{C}_{3}{+}^{51}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{55}{C}_{3}{+}^{54}{C}_{3}{+}^{53}{C}_{3}{+}^{52}{C}_{3}{+}^{52}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{55}{C}_{3}{+}^{54}{C}_{3}{+}^{53}{C}_{3}{+}^{53}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{55}{C}_{3}{+}^{54}{C}_{3}{+}^{54}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{55}{C}_{3}{+}^{55}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{56}{C}_{4}\phantom{\rule{0ex}{0ex}}$

#### 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.

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

#### Question 7:

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

We know that two lines are required for one point of intersection.
∴ Number of points of intersection =${}^{8}{C}_{2}=\frac{8}{2}×\frac{7}{1}=28$

#### 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.

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 =

#### 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.

4 white and 5 red balls are to be selected from 8 white and 10 red balls.
∴ Required number of ways =${}^{8}{C}_{4}{×}^{10}{C}_{5}$

#### Question 10:

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

Number of groups in which 12 boys are to be divided = 3
Now, 4 boys can be chosen out of 12 boys in $\left({C}_{4}{×}^{8}{C}_{4}{×}^{4}{{C}^{12}}_{4}\right)$ ways.
These groups can be arranged in 3! ways.
∴ Total number of ways =

#### Question 11:

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

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

#### Question 1:

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

(i) We have,
[∵ ]
$⇒$                             [∵ ]

(ii) We have,
${}^{12}{C}_{10}{=}^{12}{C}_{2}$                                                            [∵ ]

[∵ ]

[∵ ]

(iii) We have,

35C35 = 35C0                                                     [∵ ]

$⇒$35C35 =1                                                          [∵ ]

(iv) We have,

n + 1Cn =n + 1C1                                               [∵ ]
[∵ ]
$⇒$   n + 1Cn = n+1                                                  [∵ ]

(v) We have,

[∵ ]

[∵ ]

[∵ ]

#### Question 2:

If nC12 = nC5, find the value of n.

We have,
nC12 = nC5
[∵    or, ]

#### Question 3:

If nC4 = nC6, find 12Cn.

We have,

[∵    or, ]

Now,                                           [∵ ]
[∵ ]
[∵ ]

#### Question 4:

If nC10 = nC12, find 23Cn.

Given: nC10 = nC12
We have,

[∵    or, ]

Now,                                       [∵ ]
[∵ ]
[∵ ]

#### Question 5:

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

Given:
24Cx = 24C2x + 3
We have,
[∵  or, ]

#### Question 6:

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

Given:
18Cx = 18Cx + 2

By using   or  we get,

#### Question 7:

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

Given:
15C3r = 15Cr + 3
[∵  or, ]

#### Question 8:

If 8Cr7C3 = 7C2, find r.

Given:
8Cr7C3 = 7C2
We have,

[∵   ; $r\le n$]
[∵  or, ]

#### Question 9:

If 15Cr : 15Cr − 1 = 11 : 5, find r.

Given:
15Cr : 15Cr − 1 = 11 : 5

We have,

[∵ ]

#### Question 10:

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

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

#### Question 11:

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

We have,  28C2r : 24C2r − 4 = 225 : 11

#### Question 12:

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

Since nC4 , nC5 and nC6 are in AP.

∴ 2. nC5 = nC4 + nC6

#### Question 13:

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

Given:

or,

Now,
But, this is not possible.
∴

#### Question 14:

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

Given:

[∵ Property 5:    or   ]

Now,
[∵ ]
[∵  ]
[∵ ]

#### Question 15:

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

[∵ ]

[∵ ]

#### Question 16:

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

Let $2n$ negative integers be .

Then, product = ${\left(-1\right)}^{2n}\left(r\right)\left(r+1\right)\left(r+2\right),....,...\left(r+2n-1\right)$

This is divisible by $\left(2n\right)!.$

#### Question 17:

For all positive integers n, show that 2nCn + 2nCn − 1 = $\frac{1}{2}$(2n + 2Cn + 1).

∴ LHS = RHS

#### Question 18:

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

#### Question 19:

Evaluate .

Given:

[∵ ]

${=}^{22}{C}_{5}{+}^{22}{C}_{4}{+}^{23}{C}_{4}$                                               [∵]
${=}^{23}{C}_{5}{+}^{22}{C}_{4}{+}^{23}{C}_{4}$
${=}^{23}{C}_{5}{+}^{23}{C}_{4}\phantom{\rule{0ex}{0ex}}{=}^{24}{C}_{5}$                                                     [∵ ]

#### Question 20:

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

(a)

∴

(b)

∴ LHS = RHS

(c)

∴  LHS = RHS

(d)

[∵ ]
${=}^{n+2}{C}_{r}$                                                [∵ ]
= RHS
∴  LHS = RHS

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