# permutation and combination

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.

In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
A.     360    B.     480
C.     720    D.     5040
E.     None of these

Explanation:

The word ‘LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

3.

In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
A.     810    B.     1440
C.     2880    D.     50400
E.     5760

Explanation:

In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =     7!     = 2520.
2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in     5!     = 20 ways.
3!

Required number of ways = (2520 x 20) = 50400.

4.

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A.     210    B.     1050
C.     25200    D.     21400
E.     None of these

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)

=         7 x 6 x 5     x     4 x 3
3 x 2 x 1     2 x 1
= 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!
= 5 x 4 x 3 x 2 x 1
= 120.

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

5.

In how many ways can the letters of the word ‘LEADER’ be arranged?
A.     72    B.     144
C.     360    D.     720
E.     None of these