Top Permutation and Combination Aptitude Questions with Answers

Permutation & Combinations – Aptitude Questions with Answers

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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

Answer: Option D

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

Answer: Option C

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

Answer: Option C

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

Answer: Option D

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

4 x 3

6 x 5 x 4

x 4

6 x 5

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

7
9
10
None of these

Answer & Explanation
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:

Answer & Explanation
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

Answer: A) 25200

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 [∵ ⁿC_r = 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

Answer : Option B

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

Permutation and combination questions are an important part of aptitude preparation for placement tests and interviews. Regular practice with different types of problems can improve calculation speed, accuracy, and problem-solving skills. Payilagam helps learners strengthen these skills through practical, placement-focused training.