mahesh004 wrote:

A group of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different groups can be formed if two of the men refuse to server together?

a) 3510

b) 2620

c) 1404

d) 700

e) 635

E.

We need atleast 2 men and atleast 3 women. Can do this as follows:

1. 2 men + 4 women

2. 3 men + 3 men

For case 1, possibilities =

8C2 X 5C4 = 140

Subtract possibilities where 2 grumpy men are together (2C2 x 6C0) X 5C4 = 5

Case 1: 135

For case 2, poss=

8C3 X 5C3=560

Subtract poss. where 2 grumps are together (2C2 X 6C1) x 5C3 = 60

Case 2: 500

Total = 500 + 135 = 635