17.

Combinations
Exercise 17.1
1 A. Question

Evaluate the following:

14 C
3

Answer

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

∴ The value of is 364.

1 B. Question

Evaluate the following:

12 C
10

Answer

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

∴ The value of is 66.

1 C. Question

Evaluate the following:

35 C
35

Answer
We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

We know that 0! = 1

∴ The value of is 1.

1 D. Question

Evaluate the following:

Answer

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

∴ The value of is n + 1.

1 E. Question

Evaluate the following:

Answer

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)
⇒

We know that 0! = 1

∴ The value of is 31.

2. Question

If nC12 = nC5, find the value of n.

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 12≠5

So, the condition(ii) must be satisfied,

⇒ n = 12 + 5

⇒ n = 17.

∴ The value of n is 17.

3. Question

If nC4 = nC6, find 12 Cn.

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 4≠6

So, the condition(ii) must be satisfied,

⇒n=4+6

⇒ n = 10.

We need to find the value of

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

From (1)

∴ The value of is 66.

4. Question

If nC10 = nC12 , find 23 Cn.

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 10≠12

So, the condition(ii) must be satisfied,

⇒ n = 10 + 12

⇒ n = 22.

We need to find the value of

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

From (1)

∴ The value of is 23.

5. Question

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

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒ x = 2x + 3

⇒x=–3

We know that for a combination nCr,r≥0 and r should be an integer which is not satisfies here,

So, the condition (ii) must be satisfied,

⇒ 24 = x + 2x + 3

⇒ 3x = 21

⇒ x = 7.

∴ The value of x is 7.

6. Question

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

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ x≠x + 2

So, the condition(ii) must be satisfied,

⇒ 18 = x + x + 2

⇒ 2x = 16

⇒ x = 8.

∴ The value of x is 8.

7. Question

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

Answer

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒ 3r = r + 3

⇒ 2r = 3

⇒
We know that for a combination nCr,r≥0 and r should be an integer, which is not satisfies here,

So, the condition (ii) must be satisfied,

⇒ 15 = 3r + r + 3

⇒ 4r = 12

⇒ r = 4.

∴ The value of r is 4.

8. Question

If 8Cr – 7C3 = 7C2, find r.

Answer

Given:

⇒ 8Cr – 7C3 = 7C2

⇒ 8Cr = 7C2 + 7C3

We know that nCr + nCr + 1 = n + 1Cr + 1

⇒ 8Cr = 7 + 1C2 + 1

⇒ 8Cr = 8C3

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒r=3

Let us also check condition (ii),

⇒8=3+r

⇒r=5

∴ The values of ‘r’ are 3 and 5.

9. Question

If 15 Cr : 15 Cr – 1 = 11 : 5, find r.

Answer

Given:

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒
⇒

⇒ 5(16 – r) = 11r

⇒ 80 – 5r = 11r

⇒ 16r = 80

⇒r=5

∴ The value of r is 5.

10. Question

If n + 2C8 : n – 2P4 = 57 : 16, find n.

Answer

Given:

We know that ,

And also n! = n(n – 1)(n – 2)…………2.1

⇒ (n + 2)(n + 1)(n)(n – 1) = 21 × 20 × 19 × 18

Equating the corresponding terms on both sides we get,

⇒ n = 19

∴ The value of n is 19.

11. Question

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

Answer

Given:

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒
⇒

⇒ (2r)(2r – 1)(2r – 2)(2r – 3) = 11 × 12 × 13 × 14

Equating the corresponding terms on both sides we get,

⇒ 2r = 14

⇒r=7

∴ The value of r is 7.

12. Question

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

Answer

Given that nC4, nC5 and nC6 are in A.P.

We know that for three numbers a, b, c are in A.P, the following condition holds,

⇒ 2b = a + c

So,

⇒ 2nC5 = nC4 + nC6

Adding 2nC5 on both sides we get,

⇒ 4nC5 = nC4 + nC5 + nC5 + nC6

We know that nCr + nCr + 1 = n + 1Cr + 1

⇒ 4nC5 = n + 1C5 + n + 1C6

⇒ 4nC5 = n + 2C6

We know that ,

And also n! = n(n – 1)(n – 2)…………2.1

⇒ 24(n – 4) = n2 + 2n + n + 2

⇒ 24n – 96 = n2 + 3n + 2

⇒ n2 – 21n + 98 = 0

⇒ n2 – 14n – 7n + 98 = 0

⇒ n(n – 14) – 7(n – 14) = 0

⇒ (n – 7)(n – 14) = 0

⇒ n – 7 = 0 or n – 14 = 0

⇒ n = 7 or n = 14

∴ The values of n are 7 and 14.

13. Question

If 2n C3 : nC2 = 44 : 3, find n.

Answer

Given:

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒ 2 × (2n – 1) × 2 = 44

⇒ 2n – 1 = 11

⇒ 2n = 12

⇒n=6

∴ The value of n is 6.

14. Question

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

Answer

We know that if , then one of the following conditions need to be satisfied:

iii. p = q

iv. n = p + q

From the problem we can say that,

⇒ r≠r + 2

So, the condition(ii) must be satisfied,

⇒ 16 = r + r + 2

⇒ 2r = 14

⇒ r = 7.

∴ The value of r is 7.

15. Question
If α = mC2, then find the value of α C2.

Answer

⇒ α = mC2

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒ ..... (1)

We need to find α C2

From (1)

∴ The value of α C2 is .

16. Question

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

Answer

Let us assume the negative consecutive integers are –1, – 2,......, – (2n)

Let M be the product of the negative integers,

⇒ M = ( – 1).( – 2)( – 3)......( – 2n + 1).( – 2n)

⇒ M = ( – 1)2n (1.2.3.......(2n – 1).(2n))

⇒ M = 1.2.3......……(2n – 1).(2n)

We know that, n! = n(n – 1)(n – 2)…………2.1

⇒ M = (2n)!

∴ M is divisible by (2n)!.

17. Question
For all positive integers n, show that

Answer

Given that we need to prove .

Consider L.H.S:

We know that nCr + nCr + 1 = n + 1Cr + 1

⇒ 2n Cn + 2n Cn – 1 = 2n + 1Cn

We know that

And also n! = n(n – 1)(n – 2)…………2.1

= R.H.S

∴ L.H.S = R.H.S, thus proved.

18. Question

Prove that : 4n C2n : 2n Cn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.

Answer

Given that we need to prove:

4n C : 2n Cn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.

2n

Consider L.H.S:

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒
⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

19. Question

Evaluate

Answer

Given that we need to find the value of .

We know that nCr + nCr + 1 = n + 1Cr + 1

We know that

And also n! = n(n – 1)(n – 2)…………2.1

∴ The value of is 42504.

20 A. Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

Answer

Given that we need to prove

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

= R.H.S

∴ L.H.S = R.H.S, thus proved.

20 B. Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

n.n – 1Cr – 1 = (n – r + 1)(nCr – 1)

Answer

Given that we need to prove n.n – 1Cr – 1 = (n – r + 1).nCr – 1

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

= R.H.S

∴ L.H.S = R.H.S, thus proved.

20 C. Question
Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

Answer

Given that we need to prove

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

= R.H.S

∴ L.H.S = R.H.S, thus proved.

20 D. Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

nC + 2nCr – 1 + nCr – 2 = n + 2Cr

r

Answer

Given that we need to prove nCr + 2nCr – 1 + nCr – 2 = n + 2Cr

Consider L.H.S,

We know that nCr + nCr + 1 = n + 1Cr + 1

⇒ nCr + 2nCr – 1 + nCr – 2 = (nCr + nCr – 1) + (nCr – 1 + nCr – 2)

⇒ nCr + 2nCr – 1 + nCr – 2 = n + 1Cr + n + 1Cr – 1

⇒ nCr + 2nCr – 1 + nCr – 2 = n + 2Cr

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Exercise 17.2
1. Question

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

Given that we need to choose 11 players for a team out of available 15 players,

It is similar to choosing ‘r’ combinations out of ‘n’ items.

i.e., nCr ways.

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 11 players out of 15 players

⇒ N = 15 C11

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 1365 ways

∴ The total no. of ways of choosing 11 players out of 15 is 1365 ways.

2. Question

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

Answer

Given we need to find the different boat parties of 8, consisting of 5 boys and 3 girls.

The selections of girls and boys are to be made from 25 boys and 10 girls.

Let us assume the number of ways of choosing to be N.

⇒ N = (no. of ways of choosing 5 boys) × (No. of ways of choosing 3 girls)

⇒ N = (25 C5) × (10 C3)

We know that ,

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = (53130) × (120)

⇒ N = 6375600

∴ The total no. of different boat parties are 6375600 ways.

3. Question

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

Answer

Given that we need to find no. of ways of choosing 5 courses out of 9 courses if 2 courses are compulsory.

This is similar to choosing 3 subjects out of the remaining 7 subjects as 2 subjects are compulsory.

Let us assume the total no. of ways of choosing courses is N.

⇒ N = No. of ways of choosing 3 subjects out of 7 subjects.

⇒ N = 7C3

We know that ,

And also n! = (n)(n – 1)(n – 2)......2.1

⇒ N = 35

∴ The total no. of ways of choosing 5 subjects out of 9 subjects in which 2 are compulsory is 35 ways.

4. Question

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

Given that we need to choose 11 players for a team out of available 16 players,

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 11 players out of 16 players

⇒ N = 16 C11

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 4368 ways

(i) It is told that two players are always included.

It is similar to selecting 9 players out of the remaining 14 players as 2 players are already selected.

Let us assume the choosing the no. of ways be N1,

⇒ N1 = choosing 9 players out of 14 players

⇒ N1 = 14 C9

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒
⇒ N1 = 2002 ways

(ii) It is told that two players are always excluded.

It is similar to selecting 11 players out of the remaining 14 players as 2 players are already removed.

Let us assume the choosing the no. of ways be N2,

⇒ N2 = choosing 11 players out of 14 players

⇒ N2 = 14 C11

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N2 = 364 ways

∴ The required no. of ways are 4368, 2002, 364.

5 A. Question

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:

a particular professor is included

Answer

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (10 C2) × (20 C3)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that a particular is always included.

It is similar to selecting 1 professor and 3 students out of the remaining 9 professors and 20 students as 1
professor is already selected.

Let us assume the choosing the no. of ways be N1,

⇒ N1 = (choosing 1 professor out of 9 professors) × (choosing 3 students out of 20 students)

⇒ N1 = 9C1 × 20 C3

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N1 = 9 × 1140 ways

⇒ N1 = 10260 ways

5 B. Question

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:

a particular student is included

Answer

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (10 C2) × (20 C3)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that one student is always included.

It is similar to selecting 2 professors and 2 students out of remaining 10 professors and 19 students as 1
student is already selected.

Let us assume the choosing the no. of ways be N2,

⇒ N2 = (choosing 2 professors out of 10 professors) × (choosing 2 students out of 19 students)

⇒ N2 = 10 C2 × 19 C2

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒
⇒

⇒ N2 = 45 × 171

⇒ N2 = 7695 ways

5 C. Question

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:

a particular student is excluded.

Answer

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (10 C2) × (20 C3)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that one student is always excluded.

It is similar to selecting 2 professors and 3 students out of remaining 10 professors and 19 students as 1
student are already removed.

Let us assume the choosing the no. of ways be N3,

⇒ N3 = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 19 students)

⇒ N3 = 10 C2 × 19 C3

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N3 = 45 × 969

⇒ N3 = 43605 ways.
∴ The required no. of ways are 51300, 10260, 7695, 43605.

6. Question

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

Answer

Given that we need to find the no. of ways of obtaining a product by multiplying two or more from the
numbers 3, 5, 7, 11.

The following are the cases the product can be done,

i. Multiplying two numbers

ii. Multiplying three numbers

iii. Multiplying four numbers

Let us assume the total numbers of ways of the product be N

⇒ N = (no. of ways of multiplying two numbers) + (no. of ways of multiplying three numbers) + (no. of
multiplying four numbers)

⇒ N = 4C2 + 4C3 + 4C4

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒N=6+4+1

⇒ N = 11

∴ The total number of ways of product is 11 ways.

7. Question

From a class of 12 boys and 10 girls, 10 students are to be chosen for the 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

Given that 10 students need to be selected from 12 boys and 10 girls including at least 4 boys and 4 girls.

It is also told that the two girls must be includes who won prizes last year.

The cases that satisfy these conditions are:

i. Selecting 6 boys and 4 girls (in which 2 girls are already included)

ii. Selecting 5 boys and 5 girls (in which 2 girls are already included)

iii. Selecting 4 boys and 6 girls. (in which 2 girls are already included)

Let us assume the total no. of ways of selection be N,

⇒ N = (no. of ways of selecting 6 boys and 2 girls from remaining 12 boys and 8 girls) + (no. of ways of
selecting 5 boys and 3 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 4 boys and 4 girls
from remaining 12 boys and 8 girls)

Since, two girls are already selected,

⇒ N = (12 C6 × 8C2) + (12 C5 × 8C3) + (12 C4 × 8C4)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = (924 × 28) + (792 × 56) + (495 × 70)

⇒ N = 25872 + 44352 + 34650

⇒ N = 104874

∴ The total number of ways of product is 11 ways.

8 A. Question

How many different selections of 4 books can be made from 10 different books, if

there is no restriction

Answer

Given that we need to choose 4 books out of available 10 different books,

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 4 books out of 10 books

⇒ N = 10 C4

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 210 ways

8 B. Question

How many different selections of 4 books can be made from 10 different books, if

two particular books are always selected

Answer

It is told that two books are always selected.

It is similar to selecting 2 books out of the remaining 8 books as 2 books are already selected.
Let us assume the choosing the no. of ways be N1,

⇒ N1 = choosing 2 books out of 8 books

⇒ N1 = 8C2

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N1 = 28 ways

8 C. Question

How many different selections of 4 books can be made from 10 different books, if

two particular books are never selected.

Answer

It is told that two books are never selected.

It is similar to selecting 4 books out of remaining 8books as 2 books are already removed.

Let us assume the choosing the no. of ways be N2,

⇒ N2 = choosing 4 books out of 8 books

⇒ N2 = 8C4

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N2 = 70 ways

∴ The required no. of ways are 210, 28, 70.

9. Question

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

Given that we have 4 officers and 8 jawans, we need to choose 6 persons with the following conditions,

i. To include exactly one officer:

ii. To include at least one officer.

(i) It is told that we need to choose 6 persons with exactly one officer.

Let us assume the no. of ways of choosing to be N.

⇒ N = (no. of ways of choosing 1 officer and 5 jawans from 4 officers and 8 jawans)

⇒ N = (no. of ways of choosing 1 officer from 4 officers) × (no. of ways of choosing 5 jawans from 8 jawans)

⇒ N = (4C1) × (8C5)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 4 × 56

⇒ N = 224 ways.

(ii) It is told we need to choose 6 persons with at least 1 officers.

Let us assume the total no. of ways be N1

⇒ N1 = (No. of ways of choosing 6 persons with at least one officer)

⇒ N1 = (total no. of ways of choosing 6 persons from all 12 persons) – (no. of ways of choosing 6 persons
without any officer)

⇒ N1 = 12 C6 – 8C6

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 924 – 28

⇒ N1 = 896 ways

∴ The required no. of ways are 224 and 896.

10. Question

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

Given that we need to choose a team of 11 students with at least 5 from class XI and 5 from class XII.

It is also mentioned that each class constitutes 20 students.

There are two cases of selecting a team:

i. 6 from class XI and 5 from class XII

ii. 5 from class XI and 6 from class XII

Let us assume the total no. of ways of selecting 11 students to be N.

⇒ N = no. of ways of selecting 11 students from both classes

⇒ N = (No. of ways of selecting 6 students from class XI and 5 students from class XII) + (No. of ways of
selecting 5 students from class XI and 6 students from class XII)

⇒ N = (20 C6 × 20 C5) + (20 C5 × 20 C6)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = ((38760) × (15504)) + ((15504) × (38760))

⇒ N = 600935040 + 600935040

⇒ N = 1201870080 ways

11. Question

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

Given that 10 questions are to be answered by part A and part B by choosing at least 4 from each part.

It is also mentioned that there are 6 questions in part A and 7 in part B.

There are 3 cases to answer 10 questions:

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

Let us assume the total no. of ways of answering 10 questions be N.

⇒ N = no. of ways of answering 10 questions from both parts

⇒ N = (No. of ways of answering 4 questions from part A and 6 from part B) + (No. of ways of answering 5
questions from part A and 5 questions from part B) + (No. of ways of answering 6 questions from part A and
4 from part B)

⇒ N = (6C4 × 7C6) + (6C5 × 7C5) + (6C6 × 7C4)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 105 + 126 + 35

⇒ N = 266

∴ The total no. of ways of answering 10 questions is 266 ways.

12. Question

In an examination, a student 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 a choice.

Answer

Given that student have to answer 4 questions from 5 questions.

It is also told that questions 1 and 2 are compulsory.

It is similar to answering the 2 questions out of the remaining 3 questions as 1 and 2 are compulsory.

Let us assume the no. of ways of answering the questions be N.

⇒ N = No. of ways of answering 2 questions from remaining 3 questions.

⇒ N = 3C2

We know that ,

And also n! = (n)(n – 1)......2.1

⇒N=3

∴ The no. of answering the questions is 3.

13. Question

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

Given that we need to answer 7 questions from 2 groups which consist of 6 questions each.

It is also told the candidate is permitted to answer the utmost 5 questions from any group.

The cases for answering the 7 questions:

i. 5 questions from group 1 and 2 from group 2

ii. 4 questions from group 1 and 3 from group 2

iii. 3 questions from group 1 and 4 from group 2

iv. 4 questions from group 1 and 5 from group 2

Let us assume the total no. of ways of answering 7 questions be N.

⇒ N = no. of ways of answering 7 questions from both groups

⇒ N = (No. of ways of answering 5 questions from group 1 and 2 from group 2) + (No. of ways of answering 4
questions from group 1 and 3 from group 2) + (No. of ways of answering 3 questions from group 1 and 4
from group 2) + (No. of ways of answering 2 questions from group 1 and 5 from group 2)

⇒ N = (6C5 × 6C2) + (6C4 × 6C3) + (6C3 × 6C4) + (6C2 × 6C5)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = (6 × 15) + (15 × 20) + (20 × 15) + (15 × 6)

⇒ N = 90 + 300 + 300 + 90

⇒ N = 780

∴ The total no. of ways of choosing 7 questions is 780 ways.

14. Question

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

Given that we need to find the no. of different straight lines that can be drawn from the 10 points in which 4
are collinear.

We know that 2 points are required to draw a line and the collinear points will lie on the same line, and only
one line can be drawn by joining any two points of these collinear points.

Let us assume the no. of lines formed be N,

⇒ N = (total no. of lines formed by all 10 points) – (no. of lines formed by collinear points) + 1

Here 1 is added because only 1 line can be formed by the four collinear points.

⇒ N = 10 C2 – 4C2 + 1

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 45 – 6 + 1
⇒ N = 40

∴ The total no. of ways of different lines formed are 40.

15 A. Question

Find the number of diagonals of

a hexagon

Answer

Given that we need to find the no. of diagonals of

A hexagon

We know that the hexagon has 6 vertices and each side and diagonal can be formed by joining two vertices
of a hexagon,

We know that hexagon has 6 sides,

Let us assume the no. of diagonals of the hexagon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the hexagon)

⇒ N = 6C2 – 6

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 15 – 6

⇒N=9

∴ The total no. of diagonals formed is 9.

15 B. Question

Find the number of diagonals of

a polygon of 16 sides

Answer

a polygon of 16 sides

We have given that polygon has 16 vertices and each side and diagonal can be formed by joining two
vertices of a polygon,

Let us assume the no. of diagonals of the polygon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the polygon)

⇒ N = 16 C2 – 16

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 120 – 16

⇒ N = 104

∴ The total no. of diagonals formed are 104.

16. Question

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

Answer

Given that we need to find the no. of triangles that can be drawn from the 12 points in which 5 are collinear.

We know that 3 points are required to draw a triangle and the collinear points will lie on the same line, and
no triangle can be drawn by joining any three points of these collinear points.

Let us assume the no. of triangles formed be N,

⇒ N = (total no. of triangles formed by all 12 points) – (no. of triangles formed by collinear points)

⇒ N = 12 C3 – 5C3

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 220 – 10

⇒ N = 210

∴ The total no. of triangles formed are 210.

17. Question

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

Given that we need to find no. of ways of selecting 5 persons out of 6 men and 4 women in which at least
one woman is necessary.

Let us assume the no. of ways of selection be N.

⇒ N = (total no. of ways of selecting 5 persons out of all 10 persons) – (No. of ways of selecting 5 persons
without any women)

⇒ N = (10 C5) – (6C5)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 252 – 6

⇒ N = 246

∴ The total no. of choosing 5 persons with at least one woman is 246.

18. Question

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

Given that 52 families out of 82 families have at most 2 children.

It is told that 20 families need to be selected with at least 18 families having utmost 2 children.

The following are the cases:

i. 18 families having at most 2 children and 2 from other families

ii. 19 families are having at most 2 children and 1 from other families

iii. 20 families having at most 2 children

Let us assume that the no. of ways choosing to be N.

⇒ N = (Selecting 18 families having at most 2 children and 2 from other families) + (Selecting 19 families
having at most 2 children and 1 from other families) + (Selecting 20 families having at most 2 children)

⇒ N = (52 C18 × 35 C2) + (52 C19 × 35 C1) + (52 C20 )

We know that ,

And also n! = (n)(n – 1)......2.1

∴ The no. of ways of choosing 18 families are .

19. Question

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? (ii) at least 3 girls?

Answer

Given that a group consists of 4 girls and 7 boys.

We need to select a team of 5 members with the following conditions:

i. If a team has no girl

ii. If a team has at least one boy and one girl

iii. If a team has at least 3 girls.

i. Given that we need to select a team of 5 members with no girl present in it out of 4 girls and 7 boys.

Let us assume the no. of ways of selection be N

⇒ N = (selecting 5 members out of 7 boys without any girl)

⇒ N = 7C5

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 21 ways

The no. of ways of selecting 5 members without a girl is 21 ways.

ii. Given that we need to select team of 5 members with at least 1 boy and 1 girl.

Let us assume the no. of ways of selection be N1.

⇒ N1 = (Total ways of selecting 5 members out of all 11 members) – (No. of ways of selecting 5 members
without any girl)

⇒ N1 = (11 C5)–(7C5)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 462–21

⇒ N1 = 441

The no. of ways of selecting 5 members with at least 1 girl and 1 boy is 441.

iii. Given that we need to find the no. of ways to select 5 members with at least 3 girls out of 7 boys and 4
girls.

The following are the possible cases:

i. Selecting 3 girls and 2 boys

ii. Selecting 4 girls and 1 boy

Let us assume the no. of ways of selection be N2.

⇒ N2 = (No. of ways of selecting 3 girls and 2 boys out of 7 boys and 4 girls) + (No. of ways of selecting 4
girls and 1 boy out of 7 boys and 4 girls)

⇒ N2 = ((4C3) × (7C2)) + ((4C4) × (7C1))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N2 = (4 × 21) + (1 × 7)

⇒ N2 = 84 + 7

⇒ N2 = 91

The no. of ways of selecting 5 members with at least 3 girls is 91.

20. Question

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 this committee would consist of 1 man and 2 women?

Answer

Given that we need to select 3 persons out of 2 men and 3 women,

Let us assume the no. of ways of selecting be N,

⇒ N = selecting 3 persons out of total 5 persons

⇒ N = 5C3

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒ N = 10 ways

The no. of ways of selecting 3 persons out of 2 men and 3 women is 10.

(ii) It is told that 1 man and 2 women should be selected out of 2 men and 3 women.

Let us assume the no. of ways of selection be N1,

⇒ N1 = (selecting one man out of 2 men) × (selecting 2 women out of 3 women)

⇒ N1 = (2C1) × (3C2)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒ N1 = 6 ways

The no. of ways of selecting 1 man and 2 women is 6.

21. Question

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

Answer

i. We know that the decagon has 10 vertices and each side and diagonal can be formed by joining two
vertices of a hexagon,

We know that decagon has 10 sides,

Let us assume the no. of diagonals of the hexagon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the hexagon)

⇒ N = 10 C2 – 10

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 45 – 10

⇒ N = 35

∴ The total no. of diagonals formed is 35.

ii. Given that we need to find the no. of triangles that can be drawn in a decagon.

We know that 3 points are required to draw a triangle.

We know that decagon has 10 sides

Let us assume the no. of triangles formed be N1,

⇒ N1 = (total no. of triangles formed by all 10 points)

⇒ N1 = 10 C3

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N1 = 120

∴ The total no. of ways of different lines formed are 120.

22. Question

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

Given that we need to draw 5 cards from a deck of 52 cards.

We need to find the no. of ways that at least one of the 5 cards has to be a king.

We know that there are 4 kings present in a deck.

Let us assume the no. of ways of drawing cards be N.

⇒ N = (Total no. of ways of drawing 5 cards out of 52 cards) – (No. of ways of drawing 5 cards without a king
from remaining 48 cards)

⇒ N = (52 C5) – (48 C5)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 2598960 – 1712304

⇒ N = 886656

∴ The no. of ways of drawing 5 cards with at least 1 king is 886656.

23. Question

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

Given that we need to select 6 persons out of 8 persons, it is also mentioned that person B must be selected
in the case of selection of A.

We need to find the no. of ways of selecting 6 persons.

The possible cases for the selection of 6 persons are:

i. selecting person A and person B and 4 others(similar to selecting 4 persons out of remaining 6 persons).

ii. selecting 6 persons other than person A and person B

Let us assume the no. of ways of selection be N.

N = (Selecting 4 persons from remaining 6 persons) + (Selecting 6 persons leaving Person A and B)

N = (6C4) + (6C6)

We know that ,
And also n! = (n)(n – 1)......2.1

⇒ N = 15 + 1

⇒ N = 16

∴ The no. of ways of selecting 6 persons is 16.

24. Question

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

Answer

Given that we need to select a team consisting of 3 boys and 3 girls out of 5 boys and 4 girls.

Let us assume the no. of ways of selection be N.

⇒ N = (Selecting 3 boys out of 5 boys) × (selecting 3 girls out of 4 girls)

⇒ N = (5C3) × (4C3)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = (10) × (4)

⇒ N = 40

∴ The no. of ways of selecting 3 boys and 3 girls is 40.

25. Question

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 color.

Answer

Given that we need to select the 3 red, 3 white, and 3 blue balls out of 6 red, 5 white and 5 blue balls.

Let us assume the no. of ways of selection be N.

⇒ N = (no. of ways of selection of 3 red balls out of 6 red balls) × (no. of ways of selection of 3 white balls out
of 5 white balls) × (no. of ways of selection of 3 blue balls out of 5 blue balls)

⇒ N = (6C3) × (5C3) × (5C3)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 20 × 10 × 10

⇒ N = 2000

∴ The no. of ways of selection is 2000.

26. Question

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

Answer

Given that 5 cards are drawn out of 52 cards. We know that there are 4 aces present in a deck of 52 cards.

We need to find the no. of ways of drawing 5 cards with exactly one ace out of a deck of 52 cards.

Let us assume the no. of ways of drawing cards be N.

⇒ N = (Drawing 1 ace out of 4 aces) × (Drawing 4 cards out of remaining 48 cards)

⇒ N = (4C1) × (48 C4)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 4 × 194580

⇒ N = 778320

∴ The no. of ways of drawing 5 cards with exactly one ace is 778320.

27. Question

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

Given that we need to select 11 players out of available 17 players in which 5 players are bowlers.

It is also mentioned that each team should contain exactly 4 bowlers.

Let us assume the no. of ways of selection be N.

⇒ N = (no. of ways of selecting 4 bowlers out of 5) × (no. of ways of selecting 7 players from remaining 12
players)

⇒ N = (5C4) × (12 C7)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 5 × 792

⇒ N = 3960

28. Question

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

Given that we need to draw 2 black and 3 red balls from a bag of 5 black and 6 red balls.

Let us assume the no. of ways of drawing be N.

⇒ N = (no. of ways of selecting 2 black balls from 5 black balls) × (no. of ways of selecting 3 red balls from 6
red balls)

⇒ N = (5C2) × (6C3)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 10 × 20

⇒ N = 200

∴ The no. of ways of drawing 2 black and 3 red balls is 200.

29. Question

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

Given that a student needs to choose 5 courses out of 9 courses. It is also told 2 courses are compulsory to
the student.

It is similar to choosing 3 courses of remaining 7 courses since 2 courses are already chosen.

Let us assume the no. of ways of choosing to be N,

⇒ N = (no. of ways of choosing 3 courses out of remaining 7 courses)

⇒ N = 7C3

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N = 35

∴ The no. of ways of choosing 5 courses is 35.

30 A. Question

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 :

exactly 3 girls?

Answer

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

i. It is told we need to select 7 members out of 9 boys and 4 girls with exactly 3 girls.

Let us assume the no. of ways of selecting is N.

⇒ N = (no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)

⇒ N = (4C3) × (9C4)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 4 × 126

⇒ N = 504

The no. of ways of selecting 7 members with exactly 3 girls is 504.

30 B. Question

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 :

at least 3 girls?

Answer

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

It is told we need to select 7 members out of 9 boys and 4 girls with at least 3 girls.

The possible cases are the following:

i. Selecting 3 girls and 4 boys

ii. Selecting 4 girls and 3 boys

Let us assume the no. of ways of selecting is N1.

⇒ N1 = ((no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)) ×
((no. of ways of selecting 4 girls out of 4 girls) × (no. of ways of selecting 3 boys out of 9 boys))

⇒ N1 = ((4C3) × (9C4)) + ((4C4) × (9C3))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = (4 × 126) + (1 × 84)

⇒ N1 = 504 + 84

⇒ N1 = 588

The no. of ways of selecting 7 members with at least 3 girls is 588.

30 C. Question

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 :

at most 3 girls?

Answer

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

It is told we need to select 7 members out of 9 boys and 4 girls with at most 3 girls.

The possible cases are the following:

i. Selecting 3 girls and 4 boys

ii. Selecting 2 girls and 5 boys

iii. Selecting 1 girl and 6 boys

iv. Selecting 7 boys

Let us assume the no. of ways of selecting is N2.

⇒ N2 = ((no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)) ×
((no. of ways of selecting 2 girls out of 4 girls) × (no. of ways of selecting 5 boys out of 9 boys)) × ((no. of
ways of selecting 1 girls out of 4 girls) × (no. of ways of selecting 6 boys out of 9 boys))

⇒ N2 = ((4C3) × (9C4)) + ((4C2) × (9C5)) + ((4C1) × (9C6)) + (9C7)

We know that ,
And also n! = (n)(n – 1)......2.1

⇒ N2 = (4 × 126) + (6 × 126) + (4 × 84) + (36)

⇒ N2 = 504 + 756 + 336 + 36

⇒ N2 = 1632

The no. of ways of selecting 7 members with at most 3 girls is 1632.

31. Question

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 help a student select the questions?

Answer

Given that a student needs to answer 8 questions out of 12 questions in which 5 from part I and 7 from part
II.

It is also told that student needs to answer at least 3 questions from each part.

The possible cases are the following:

i. 3 from part I and 5 from part II

ii. 4 from part I and 4 from part II

iii. 5 from part I and 3 from part II

Let us assume the no. of ways of answering be N.

⇒ N = (no. of ways of answering 3 questions from part I and 5 from part II) + (no. of ways of answering 4
questions from part I and 4 from part II) + (no. of ways of answering 5 questions from part I and 3 from part
II)

⇒ N = ((5C3) × (7C5)) + ((5C4) × (7C4)) + ((5C5) × (7C3))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = (10 × 21) + (5 × 35) + (35)

⇒ N = 420

∴ The no. of ways of answering the question paper is 420.

32. Question

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

Answer

We know that parallelogram has 4 lines in which 2 sides are parallel to each other which means 2 pairs of
lines are parallel lines

It is told that parallelogram is cut by two sets of m lines parallel to its sides.

This means there will be two sets of (m + 2) lines parallel to each other.

We need two sets of parallel lines to form a parallelogram in which the lines need to be chosen from these
two sets of (m + 2) parallel lines.

Let us assume that the no. of parallelograms formed be N.

⇒ N = (choosing 2 parallel lines from (m + 2) parallel lines which are parallel to one side) × (choosing 2
parallel lines from (m + 2) parallel lines which are parallel to the side which is not parallel to the first side)

⇒ N = (m + 2C2) × (m + 2C2)

⇒ N = (m + 2C2)2

We know that ,

And also n! = (n)(n – 1)......2.1

∴ The total no. of parallelograms formed are .

33. Question

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

Given that we need to find the no. of different straight lines that can be drawn from the 18 points in which 5
are collinear.

We know that 2 points are required to draw a line and the collinear points will lie on the same line, and only
one line can be drawn by joining any two points of these collinear points.

Let us assume the no. of lines formed be N,

⇒ N = (total no. of lines formed by all 18 points) – (no. of lines formed by collinear points) + 1

Here 1 is added because only 1 line can be formed by the four collinear points.

⇒ N = 18 C2 – 5C2 + 1

We know that ,
And also n! = (n)(n – 1)......2.1

⇒ N = 153 – 10 + 1

⇒ N = 144

∴ The total no. of ways of different lines formed are 144.

Given that we need to find the no. of triangles that can be drawn from the 18 points in which 5 are collinear.

We know that 3 points are required to draw a triangle and the collinear points will lie on the same line, and
no triangle can be drawn by joining any three points of these collinear points.

Let us assume the no. of triangles formed be N,

⇒ N = (total no. of triangles formed by all 18 points) – (no. of triangles formed by collinear points)

⇒ N = 18 C3 – 5C3

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N = 816 – 10

⇒ N = 806

∴ The total no. of triangles formed are 806.

Exercise 17.3
1. Question

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

Answer

Given that we need to find the no. of words formed by 2 vowels and 3 consonants which were taken from 5
vowels and 17 consonants.

Let us find the no. of ways of choosing 2 vowels and 3 consonants and assume it to be N1.

⇒ N1 = (No. of ways of choosing 2 vowels from 5 vowels) × (No. of ways of choosing 3 consonants from 17
consonants)

⇒ N1 = (5C2) × (17 C3)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒ N1 = 10 × 680

⇒ N1 = 6800

Now we need to find the no. of words that can be formed by 2 vowels and 3 consonants.

Now we need to arrange the chosen 5 letters. Since every letter differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the
total no. of words that can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N1 × 5!

⇒ N = 6800 × 120

⇒ N = 816000

∴ The no. of words that can be formed containing 2 vowels and 3 consonants are 816000.

2. Question

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

Answer

Given that 5 persons need to be selected from 10 person P1, P2, P3,……P10 .

It is also told that P1 should be present and P4 and P5 should not be present.

It is similar to choosing 4 persons from remaining 7 persons as P1 is selected and P 4 and P5 are already
removed.

Let us first find the no. of ways to choose persons and assume it to be N1.

⇒ N1 = Selecting 4 persons from remaining 7 persons

⇒ N1 = 7C4

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 35

Now we need to arrange the chosen 5 people. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the
persons can be arranged in 5! Ways.

Let us assume the total possible arrangements be N.

⇒ N = N1 × 5!

⇒ N = 35 × 120

⇒ N = 4200

∴ The total no. of possible arrangement can be done is 4200.

3 A. Question

How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming
that no letter is repeated, if

4 letters are used at a time

Answer

Given that we need to find the no. of words formed by 4 letters which were taken from word ‘MONDAY.’

Let us find the no. of ways of choosing 4 letters and assume it to be N1.

⇒ N1 = (No. of ways of choosing 4 letters from MONDAY)

⇒ N1 = (6C4)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 15

Now we need to find the no. of words that can be formed by 4 letters.

Now we need to arrange the chosen 4 letters. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the
total no. of words that can be formed is 4!.

Let us the total no. of words formed be N.

⇒ N = N1 × 4!

⇒ N = 15 × 24

⇒ N = 360

∴ The no. of words that can be formed by 4 letters of MONDAY is 360.

3 B. Question

How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming
that no letter is repeated, if

all letters are used at a time

Answer

Given that we need to find the no. of words formed by all letters of MONDAY.

Now we need to arrange the 6 letters. Since every letter differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the
total no. of words that can be formed is 6!.
Let us the total no. of words formed be N.

⇒ N = 6!

⇒ N = 360

∴ The no. of words that can be formed by 6 letters of MONDAY is 360.

3 C. Question

How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming
that no letter is repeated, if

all letters are used but the first letter is a vowel?

Answer

Given that we need to find the no. of words formed by all letters from MONDAY in which the first letter should
be a vowel.

In MONDAY the vowels are O and A. We need to choose one vowel from these 2 vowels for the first place of
the word.

Let us find the no. of ways of choosing vowel and assume it to be N1.

⇒ N1 = (No. of ways of choosing a vowel from 2 vowels)

⇒ N1 = (2C1)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 2

Now we need to find the no. of words that can be formed by remaining 5 letters.

Now we need to arrange the remaining 5 letters. Since every letter differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N1 × 5!

⇒ N = 2 × 120

⇒ N = 240

∴ The no. of words that can be formed by all letters of MONDAY in which the first letter is a vowel is 240.

4. Question

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

Answer

Given that we need to find the no. of permutations formed by r things which were taken from n distinct
things in which 3 particular things must occur together.

Here, it is clear that 3 things are already selected and we need to choose (r – 3) things from the remaining (n
– 3) things.
Let us find the no. of ways of choosing (r – 3) things and assume it to be N1.

⇒ N1 = (No. of ways of choosing (r – 3) things from remaining (n – 3) things)

⇒ N1 = n – 3Cr – 3

Now we need to find the no. of permutations than can be formed using 3 things which are together.

Now we need to arrange the chosen 3 things. Since every thing differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is 3!.

Now let us assume the together things as a single thing this gives us total (r – 2) things which were present
now.

Now, we need to arrange these (r – 2) things. Since every thing differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is (r – 2)!.

Let us the total no. of words formed be N.

⇒ N = N1 × 3! × (r – 2)!

⇒ N = n – 3Cr – 3 × 3! × (r – 2)!

∴ The no. of permutations that can be formed by r things which are chosen from n things in which 3 things
are always together is n – 3Cr – 3 × 3! × (r – 2)!.

5. Question

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

Answer

Given the word is INVOLUTE. We have 4 vowels namely I,O,U,E, and consonants namely N,V,L,T.

We need to find the no. of words that can be formed using 3 vowels and 2 consonants which were chosen
from the letters of involute.

Let us find the no. of ways of choosing 3 vowels and 2 consonants and assume it to be N1.

⇒ N1 = (No. of ways of choosing 3 vowels from 4 vowels) × (No. of ways of choosing 2 consonants from 4
consonants)

⇒ N1 = (4C3) × (4C2)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 4 × 6

⇒ N1 = 24

Now we need to find the no. of words that can be formed by 3 vowels and 2 consonants.

Now we need to arrange the chosen 5 letters. Since every letter differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N1 × 5!

⇒ N = 24 × 120

⇒ N = 2880

∴ The no. of words that can be formed containing 3 vowels and 2 consonants chosen from INVOLUTE is 2880.

6. Question

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

Answer

Given that we need to find the no. of permutations formed by r things which were taken from n distinct
things in which 2 specified things must occur together.

Here, it is clear that 2 things are already selected and we need to choose (r – 2) things from the remaining (n
– 2) things.

Let us find the no. of ways of choosing (r – 2) things and assume it to be N1.

⇒ N1 = (No. of ways of choosing (r – 2) things from remaining (n – 2) things)

⇒ N1 = n – 2Cr – 2

Now we need to find the no. of permutations than can be formed using 2 things which are together.

Now we need to arrange the chosen 2 things. Since every thing differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is 2!.

Now let us assume the together things as a single thing this gives us total (r – 1) things which were present
now.

Now, we need to arrange these (r – 1) things. Since every thing differs from other. The arrangement is
similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that
can be formed is (r – 1)!.

Let us the total no. of words formed be N.

⇒ N = N1 × 2! × (r – 1)!

⇒ N = n – 2Cr – 2 × 2 × (r – 1)!

∴ The no. of permutations that can be formed by r things which are chosen from n things in which 3 things
are always together are n – 2Cr – 2 × 2 × (r – 1)!.

7. Question

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

Given the word is PROPORTION. The letters present in it are:

P: 2 in number

R: 2 in number

O: 3 in number

T: 1 in number

I: 1 in number

N: 1 in number

a. We need to find the no. of ways of selecting 4 letters from the word proportion:
The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. There are 6 different letters from which we need to select 4 letters. Let us assume no. of ways of selection
be N1

⇒ N1 = no. of ways of selecting 4 letters from 6 letters

⇒ N1 = 6C4

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 15

ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 3 and 2
distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N2

⇒ N2 = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2
distinct letters from remaining 5 distinct letters)

⇒ N2 = (3C1) × (5C2)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N2 = 3 × 10

⇒ N2 = 30

iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of
ways of selection be N3

⇒ N3 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N3 = 3C2

We know that ,

And also n! = (n)(n – 1)......2.1

⇒
⇒

⇒ N3 = 3

iv. There is only 1 letter which occurred thrice, and 1 letter needs to be selected from the remaining 5
distinct letters.

Let us assume no. of ways of selection be N4

⇒ N4 = no. of ways of selecting 1 letter from 5 letters

⇒ N4 = 5C1

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N4 = 5

Total no. of ways of selection = N1 + N2 + N3 + N4

Total no. of ways of selection = 15 + 30 + 3 + 5

Total no. of ways of selection = 53

b. We need to find the no. of ways of arranging 4 letters from the word proportion:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. Now we need to arrange the chosen 4 different letters. Since every word differs from other. The
arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total
no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N5.

⇒ N5 = N1 × 4!

⇒ N5 = 15 × 24

⇒ N5 = 360

ii. Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to
that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of

arrangements that can be made are .

Let us assume no. of ways of arrangement be N6.

⇒ N6 = N2 ×
⇒ N6 = 30 × 4 × 3

⇒ N6 = 360

iii. Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The
arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are
similar of another type which are ways too arrange. So, the total no. of arrangements that can be made

are .

Let us assume no. of ways of arrangement be N7.

⇒ N7 = N3 ×

⇒ N7 = 18

iv. Now we need to arrange the chosen 1 different letters and 3 alike letters. The arrangement is similar to
that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of

arrangements that can be made are .

Let us assume no. of ways of arrangement be N8.

⇒ N8 = N4 ×

⇒ N8 = 5 × 4

⇒ N8 = 20

Total no. of ways of arrangement = N5 + N6 + N7 + N8

Total no. of ways of arrangement = 360 + 360 + 18 + 20

Total no. of ways of arrangement = 758

8. Question

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

Answer

Given the word is MORADABAD. The letters present in it are:

M: 1 in number

O: 1 in number

R: 1 in number

A: 3 in number

D: 2 in number

B: 1 in number

a. We need to find the no. of words formed by 4 letters from the word MORADABAD:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. There are 6 different letters from which we need to select 4 letters. Let us assume no. of ways of selection
be N1

⇒ N1 = no. of ways of selecting 4 letters from 6 letters

⇒ N1 = 6C4

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 15

Now we need to arrange the chosen 4 different letters. Since every word differs from other. The
arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total
no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N2.

⇒ N2 = N1 × 4!

⇒ N2 = 15 × 24

⇒ N2 = 360

ii. There are 2 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2
distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N3

⇒ N3 = (no. of ways of selecting 2 alike letters from the 2 types of alike letters) × (no. of ways of selecting 2
distinct letters from remaining 5 distinct letters)

⇒ N3 = (2C1) × (5C2)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N3 = 2 × 10

⇒ N3 = 20

Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that
of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of

arrangements that can be made are .

Let us assume no. of ways of arrangement be N4.

⇒ N4 = N3 ×

⇒ N4 = 20 × 4 × 3
⇒ N4 = 240

iii. There are 2 letters which occurred more than once from which we need to select 2. Let us assume no. of
ways of selection be N5

⇒ N5 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N5 = 2C2

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N5 = 1

Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The
arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are
similar of another type which are ways too arrange. So, the total no. of arrangements that can be made

are .

Let us assume no. of ways of arrangement be N6.

⇒ N6 = N5 ×

⇒ N6 = 6

iv. There is only 1 letter which occurred thrice, and 1 letter needs to be selected from the remaining 5
distinct letters.

Let us assume no. of ways of selection be N7

⇒ N7 = no. of ways of selecting 1 letter from 5 letters

⇒ N7 = 5C1

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N7 = 5

Now we need to arrange the chosen 1 different letters and 3 alike letters. The arrangement is similar to that
of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of

arrangements that can be made are .

Let us assume no. of ways of arrangements be N8.

⇒ N8 = N7 ×

⇒ N8 = 5 × 4

⇒ N8 = 20

Total no. of ways of words formed = N2 + N4 + N6 + N8

Total no. of ways of words formed = 360 + 240 + 6 + 20

Total no. of ways of words formed = 626

9. Question

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

Answer

Given that we need to accommodate 21 guests to two round tables which can accommodate 15 and 6
persons each.

We need to select 6 members first and arrange 6 and 15 members accordingly in the respective tables.

Let us assume the no. of ways of choosing 6 members to be N1

⇒ N1 = No. of ways of choosing 6 members out of 21 guests.

⇒ N1 = 21 C6

Now we need to arrange the 6 members in a round table. By fixing a guest at a single seat, We arrange the
remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n!
ways to arrange. So, the total no. of arrangements that can be made is 5!.

Now we need to arrange the 15 members in a round table. By fixing a guest at a single seat, We arrange the
remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n!
ways to arrange. So, the total no. of arrangements that can be made is 14!.

Let us assume total no. of ways of arranging the guests in the table be N

⇒ N = N1 × 5! × 14!

⇒ N = 21 C6 × 5! × 14!

∴ The no. of ways of accommodating guests is 21 C6 × 5! × 14!.

10. Question

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

Answer

Given the word is EXAMINATION. The letters present in it are:

E: 1 in number

X: 1 in number

A: 2 in number

M: 1 in number

I: 2 in number

N: 2 in number

T: 1 in number

O: I in number

a. We need to find the no. of words formed by 4 letters from the word EXAMINATION:
The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

i. There are 8 different letters from which we need to select 4 letters. Let us assume no. of ways of selection
be N1

⇒ N1 = no. of ways of selecting 4 letters from 8 letters

⇒ N1 = 8C4

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 70

Now we need to arrange the chosen 4 different letters. Since every word differs from other. The
arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total
no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N2.

⇒ N2 = N1 × 4!

⇒ N2 = 70 × 24

⇒ N2 = 1680

ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2
distinct letters from the remaining 7 distinct letters. Let us assume no. of ways of selection be N3

⇒ N3 = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2
distinct letters from remaining 5 distinct letters)

⇒ N3 = (3C1) × (7C2)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N3 = 3 × 21

⇒ N3 = 63

Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that
of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of
arrangements that can be made are .

Let us assume no. of ways of arrangement be N4.

⇒ N4 = N3 ×

⇒ N4 = 63 × 4 × 3

⇒ N4 = 756

iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of
ways of selection be N5

⇒ N5 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of other type

⇒ N5 = 3C2

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N5 = 3

Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The
arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are
similar of another type which are ways too arrange. So, the total no. of arrangements that can be made

are .

Let us assume no. of ways of arrangement be N6.

⇒ N6 = N5 ×

⇒ N6 = 18

Total no. of combinations = N1 + N3 + N5

Total no. of combinations = 70 + 63 + 3

Total no. of combinations = 136

Total no. of ways of permutations = N2 + N4 + N6

Total no. of ways of permutations = 1680 + 756 + 18

Total no. of ways of permutations = 2454

11. Question

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 are seated?

Answer

Given that 16 persons need to be seated along two sides of a long table with 8 persons on each side.

It is also told that 4 persons sit on a particular side and 2 on the other side.
We need to choose 4 members to sit on one side from the remaining 10 members as the 6 members are
already fixed about their seatings and arrange 8 members on both sides accordingly.

Let us first find the no. of ways to choose 4 members and assume it to be N1.

⇒ N1 = Selecting 4 members from remaining 7 person members

⇒ N1 = 10 C4

We know that ,

And also n! = (n)(n – 1)......2.1

⇒ N1 = 210

Now we need to arrange the chosen 8 members. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the
persons can be arranged in 8! Ways.

This will be the same for both the tables.

Let us assume the total possible arrangements be N.

⇒ N = N1 × 8! × 8!

⇒ N = 210 × 8! × 8!

∴ The total no. of ways of seating arrangements can be done 210 × 8! × 8!.

Very Short Answer

1. Question

Write in the simplified form.

Answer

We know:

nC + nCr-1 = n+1Cr ⇒ (1)

r

n+rC = nC0 + n+1C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1

r

n+rC = n+1C0 + n+1C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1 ⇒ (nC0 = n+1C0)

r

Using equation (1),

n+rC = n+2C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1

r

n+rC = n+3C2 + n+3C3 + . . . . . . + n+mCn+1

r

Proceeding in the same way :

n+rC = n+mCm-1 + n+mCm = n+m+1Cn+1

r
 n+rC = n+m+1Cn+1
r

2. Question

If , then write the values of n.

Answer
35 C = 35 C4n-2
n+7

n + 7 + 4n – 2 = 35 ( nCx = nCy ⇒ n = x + y or x = y )

5n + 5 = 35

5n = 30

n=6

And,

n + 7 = 4n – 2

3n = 9

n=3

3. Question

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 selection 2 out of n = nC2

Out of these lines, n lines are sides of the polygon.

Number of diagonals of the polygon =

nC –n –n
2

4. Question

Write the expression in the simplest form.

Answer

nC + nCr-1 + 2 nC
r+1 r

= (nCr+1 + nCr) + (nCr + nCr-1)

⇒ (nCr + nCr-1 = n+1Cr)

= n+1Cr+1 + n+1Cr

⇒ (nCr + nCr-1 = n+1Cr)

= n+2Cr+1

5. Question
Write the value of .

Answer

We know,

nC + nCr-1 = n+1Cr
r

Now, we have,

56-rC + 50 C4
3

= 55 C3 + 54 C3 + 53 C3 + 52 C3 + 51 C3 + 50 C3 + 50 C4

= 55 C3 + 54 C3 + 53 C3 + 52 C3 + 51 C3 + 51 C4

= 55 C3 + 54 C3 + 53 C3 + 52 C3 + 52 C4

= 55 C3 + 54 C3 + 53 C3 + 53 C4

= 55 C3 + 54 C3 + 54 C4

= 55 C3 + 55 C4

= 56 C4

6. Question

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 ways. (3 ways)

Now, out of three, if two letters have been put in the current envelope, then the last one has been put in

the correct envelope as well. This can be done in 3C3 ways. (1 way)

Number of ways = 3 + 1 = 4

Number of ways in which no letter is put in correct envelope = 6 – 4 = 2

7. Question

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 =

8C
2

= 28

8. Question

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 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.

Two parallel lines from the set of three parallel lines can be chosen in 3C2 ways.

Number of parallelograms that can be formed =

4C 3C
2 2

=6 3

= 18

9. Question

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 10 C

5

= 70 252

= 17640

10. Question

Write the number of ways in which 12 boys may be divide 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 (12 C4 8C

4
4C )
4 ways.

These groups can be arranged in 3! ways.

Total number of ways =

= 5775

11. Question

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 in 4C2 5C

3 ways.

The total number of letters is 5. These letters can be arranged in 5! ways.

Total number of words =

4C 5C 5!
2 3
=6 10 120

= 7200

MCQ
1. Question

Mark the correct alternative in the following:

If , then is equal to

A. 4896

B. 816

C. 1632

D. none of these

Answer

r + r – 10 = 20 ( nCx = nCy ⇒ n = x + y or x = y )

2r – 10 = 20

2r = 30

r = 15

Now,

18 C = 18 C15
r

18 C = 18 C3
15

18 C
3

2. Question

Mark the correct alternative in the following:

If , then is equal to

A. 54

B. 56

C. 58

D. none of these

Answer

r + r + 4 = 20 ( nCx = nCy ⇒ n = x + y or x = y )

2r + 4 = 20

2r = 16

r=8

Now,

rC = 8C3
3
8C
3

3. Question

Mark the correct alternative in the following:

If , then r is equal to

A. 5

B. 4

C. 3

D. 2

Answer

3r + r + 3 = 15 ( nCx = nCy ⇒ n = x + y or x = y )

4r + 3 = 15

4r = 12

r=3

4. Question

Mark the correct alternative in the following:

If , then r is equal to

A. 10

B. 11

C. 19

D. 12

Answer

r + 1 + r – 1 = 20 ( nCx = nCy ⇒ n = x + y or x = y )

2r = 20

r = 10

5. Question

Mark the correct alternative in the following:

If , then is equal to

A. 231

B. 210

C. 252

D. 303

Answer

nC n
12 = C8

n = 12 + 8 ( nCx = nCy ⇒ n = x + y or x = y )

n = 20
Now,

22 C = 22 C20
n

= 231

6. Question

Mark the correct alternative in the following:

If , then

A. 2 m = n

B. 2 m = n (n + 1)

C. 2 m = n (n− 1)

D. 2 n = m (m − 1)

Answer

mC = nC2
1

2m = n(n – 1)

7. Question

Mark the correct alternative in the following:

If , then n =

A. 20

B. 12

C. 6

D. 30

Answer

n = 12 + 8 ( nCx = nCy ⇒ n = x + y or x = y )

n = 20

8. Question

Mark the correct alternative in the following:

If , then x =

A. r

B. r− 1

C. n

D. r + 1
Answer

nC + nCr+1 = n+1Cx (Given)

r

Now, we have ⇒

nC + nCr-1 = n+1Cr ⇒ (1)

r

From (1) and (Given) we have,

n+1C = n+1Cx
r+1

r + 1 = x ( nCx = nCy ⇒ n = x + y or x = y )

9. Question

Mark the correct alternative in the following:

If , then a =

A. 2

B. 3

C. 4

D. none of these

Answer

a2 – a = 2 + 4 ( nCx = nCy ⇒ n = x + y or x = y )

a2 – a – 6 = 0

a2 – 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 since it’s negative.

So, a = 3.

10. Question

Mark the correct alternative in the following:

is equal to

A. 30

B. 31

C. 32

D. 33

Answer

5C + 5C2 + 5C3 + 5C4 + 5C5

1

= 5C1 + 5C2 + 5C2 + 5C1 + 5C5 (5C1 = 5C4 and 5C2 = 5C3)

=2 5C +2 5C + 5C5
1 2
=2 5+2 +1

= 10 + 20 + 1

= 31

11. Question

Mark the correct alternative in the following:

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

2 out of 4 vowels can be chosen in 4C2 ways.

3 out of 5 consonants can be chosen in 5C3 ways.

Thus, there are (4C2 5C )

3 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 5C )
3 5!

= 60 120

= 7200

12. Question

Mark the correct alternative in the following:

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

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

= 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 gives only one line.

Required number of straight lines =

66 – 3 + 1

= 64

13. Question

Mark the correct alternative in the following:

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

Three persons can take 5 seats in 5C3 ways.

Also, 3 persons can sit in 3! ways.

Required number of ways =

5C 3!
3

= 10 6

= 60

14. Question

Mark the correct alternative in the following:

In how many was 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

Required number of ways =

4C 6C + 4C2 6C + 4C3 6C + 4C4 6C
1 4 3 2 1

= 60 + 120 + 60 + 6

= 246

15. Question

Mark the correct alternative in the following:

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

Number of straight lines formed by joining the 10 points if we take 2 points at a time = 10 C2

Number of straight lines formed by joining the 4 points if we take 2 points at a time = 4C2

But, 4 collinear points, when joined in pairs give only one line.

Required number of straight lines =

45 – 6 + 1

= 40

16. Question

Mark the correct alternative in the following:

There are 13 players of cricket, out of which 4 are bowlers. In how many ways a team eleven be selected
from them so as to include at least two bowlers?

A. 72

B. 78

C. 42

D. none of these

Answer

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 9C

9 + 4C3 9C
8 + 4C4 9C
7

= 6 + 36 + 36

= 78

17. Question

Mark the correct alternative in the following:

If , then is equal to

A. 56

B. 120

C. 28

D. 91

Answer

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,

C(n,0) + C(n,1) + C(n,2) + . . . . . . . . . . + C(n,n) = 2n

Comparing the given equation with the above equation we get,

2n = 256

2n = 28

n=8

2n C = 16 C2
2

= 120

18. Question

Mark the correct alternative in the following:

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.

B.

C.

D. none of these

Answer

A host lady can invite 8 out of 12 people in 12 C8 ways.

Two out of these 12 people do not want to attend the party together.

Number of ways =

12 C - 10 C6
8

19. Question

Mark the correct alternative in the following:

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 point is

A. 216

B. 156

C. 172

D. none of these

Answer

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 = 11 C3

= 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 once circle.

 Required number of circles =

165 – 10 + 1

= 156

20. Question

Mark the correct alternative in the following:

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

Number of committees that can be formed ⇒

= 6C3 4C
2

= 120

21. Question

Mark the correct alternative in the following:

If , then the value of r is

A. 12

B. 8

C. 6

D. 10

E. 14

Answer

r – 6 + 3r + 1 = 43 ( nCx = nCy ⇒ n = x + y or x = y )

4r – 5 = 43

4r = 48

r = 12

22. Question

Mark the correct alternative in the following:

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

An octagon has 8 vertices.

The number of diagonals of a polygon is given by . (Where n = number of vertices)

Number of diagonals of an octagon =

23. Question

Mark the correct alternative in the following:

The value of is

A.

B.

C. `

D.

Answer

( 7C0 + 7C1 ) + ( 7C1 + 7C2 ) + ( 7C2 + 7C3 ) + ( 7C3 + 7C4 ) + ( 7C4 + 7C5 ) + ( 7C5 + 7C6 ) + ( 7C6 + 7C7 )

=1+2 7C +2 7C +2 7C +2 7C +2 7C +2 7C +1
1 2 3 4 5 6

=1+2 7C +2 7C +2 7C +2 7C +2 7C +2 7C + 1 ⇒ (7C3 = 7C4 and 7C2 = 7C5)

1 2 3 3 2 6

= 2 + 22 (7C1 + 7C2 + 7C3)

= 2 + 22 (7 + 6+ 5)

= 2 + 252

= 254

= 28 – 2

24. Question

Mark the correct alternative in the following:

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

5 out of 14 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 4 bowlers.

Number of ways = 5C4 9C

7 + 5C5 9C
6

= 180 + 84

= 264

25. Question

Mark the correct alternative in the following:

A lady gives a dinner party for six guests. The number of ways in which they may be selected from among
tem friends if two of the friends will not attend the party together is

A. 112

B. 140

C. 164

D. none of these

Answer

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 selecting 6 guests =8C6 = 28

If one of them attends the party, then the number of ways of selecting 6 guests = 2 8C = 112
5

Total number of ways = 112 + 28 = 140

26. Question

Mark the correct alternative in the following:

If , then n =

A. 3

B. 4

C. 5

D. 6

Answer

n+1C =2 nC
3 2

n+1=6

n=5

27. Question

Mark the correct alternative in the following:

The number of parallelogram 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

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.

Two parallel lines from the set of three parallel lines can be chosen in 3C2 ways.

Number of parallelograms that can be formed =

4C 3C
2 2

=6 3

= 18