# Permutations and Combinations Quiz Set 011

### Question 1

What is the value of 7C2?

A

21.

B

31.

C

11.

D

41.

Soln.
Ans: a

7C2 is \${7 !}/{(7 - 2) ! × 2 !}\$ = 21.

### Question 2

The letters of the word 'PERIOD' have to be arranged such that the vowels occupy only the odd positions. How many different ways are possible?

A

36.

B

46.

C

26.

D

56.

Soln.
Ans: a

This word has 6 letters, out of which 3 are consonants and 3 are vowels. The vowels have to occupy three fixed odd positions. We can place 3 vowels in first odd place, 2 in second odd place and 1 in the third odd place, giving 3 × 2 × 1 = 6 permutations. This will be done with the consonants also. So the total possibilities are 6 × 6 = 36.

### Question 3

In how many ways can 3 letter words be formed by using a given set of 6 consonants?

A

120.

B

125.

C

118.

D

121.

Soln.
Ans: a

The first place can get any of the 6 consonants, the second place can then get (6 - 1), and likewise. This is expressed as 6 × 5 × 4, which evaluates to 120.

### Question 4

The letters of the word 'HOLIDAY' have to be arranged such that the vowels come together. How many different ways are possible?

A

720.

B

730.

C

710.

D

740.

Soln.
Ans: a

This word has 7 letters, out of which 4 are consonants and 3 are vowels. The vowels have to occupy three contiguous positions. This triad can be arranged in 3! ways like this: we can place 3 vowels in first place, 2 in second place and 1 in the third place, giving 3 × 2 × 1 = 6 permutations. Next, we have to arrange the 4 consonants and the triad treated as one letter, giving 5! = 120 possibilities. So the total possibilities are 6 × 120 = 720.

### Question 5

In how many ways can 4 prizes be given to 3 students if each of them is equally eligible?

A

64.

B

74.

C

94.

D

84.

Soln.
Ans: a

The first has 4 options, the second also has 4 options, and so on. This is expressed as 4 × 4 × 4, which evaluates to 64. 