Permutation & Combinations - Aptitude Questions with Answers

Permutation & Combinations – Aptitude Questions with Answers

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

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