## Permutation & Combinations – Aptitude Questions with Answers

### This blog explains about Permutation & Combinations – Aptitude Questions with Answers and is given below :

1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
 A 564 B 645 C 735 D 756 E None of these

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)

 = 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1

 = 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
= (525 + 210 + 21)
= 756.

2. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
 A 32 B 48 C 64 D 96 E None of these

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

 = 3 x 6 x 5 + 3 x 2 x 6 + 1 2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

3 . How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
 A 40 B 400 C 5040 D 2520

Explanation:

‘LOGARITHMS’ contains 10 different letters.

 Required number of words = Number of arrangements of 10 letters, taking 4 at a time. = 10P4 = (10 x 9 x 8 x 7) = 5040.
4 . In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
 A 159 B 194 C 205 D 209 E None of these

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number
of ways
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)

 = (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 + 90 + 80 + 15)
= 209.

5 . In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is

1. 7
2. 9
3. 10
4. None of these

Let n be the number of teams.
nC2 = 21
(n(n-1)/2) = 21
⇒ n(n-1) = 42 ∴
⇒ n = 7

6. In an examination, a candidate is required to pass all five different subjects. The number of ways he can fail is:

1. 32
2. 31
3. 30
4. 29

The candidate will fail if he fails either in 1 or 2 or 3 or 4 or 5 subjects,

∴ Required number of ways 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31

7 . Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 A) 25200 B) 52000 C) 120 D) 24400

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C37C3*4C24C2)

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

 8 . There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed? A. 10420 B. 11 C. 11760 D. None of these

Explanation:

We need to select 5 men from 8 men and 6 women from 10 women

Number of ways to do this
8C5 × 10C6
8C3 × 10C4 [∵ nCr = nC(n-r)]
=(8×7×63×2×1)(10×9×8×74×3×2×1)

=56×210=11760

9 . What is the value of 100 P 2 ?

a) 990 b) 9900 c) 9990 d ) None

Explanation :

100 P 2 =  100 * 99 = 9900

10 . What is the value of 58 C 3 ?

a) 30856  b)30868  c)30654  d) None of the above

Answer : Option A ) 30856

Explanation :

58 C3 = 58 * 57 * 56 / 3 * 2 * 1

=  30856