Hi, Please help me to solve this question. Source- Princeton review A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635 The answer is (E) I tried solving by 8c1 X 8c1 X 5c3, it doesn't work because of the given constraint. Please explain me the problem step by step.

Hi, Case 1: 2 Men and 4 W # of ways of choosing 4 W out of 5 = 5C4 = 5 # of ways of choosing 2 M out of 8 (such that 2 men do not work together) = 8C2 - 1 = 28 - 1 = 27 Overall = 135 ways Case 2: 3 Men and 3 W # of ways of choosing 3 W out of 5 = 5C3 = 10 # of ways of choosing 3 M out of 8 (such that 2 men do not work together) = 8C3 - 6C1 = 56 - 6 = 50 Overall = 500 ways Hence, overall = 635 ways. Hope this helps. Thanks.

Thanks very much, I was thinking of similiar ways.. but got struck somewhere..

@4GMAT_Mumbai, can u please explain how u got 6C1 below? # of ways of choosing 3 M out of 8 (such that 2 men do not work together) = 8C3 - 6C1 = 56 - 6 = 50 Thanks

Hi Vijaynaik, # of ways of choosing 3 M out of 8 (such that 2 men do not work together) = 8C3 - 6C1 = 56 - 6 = 50 How did we get 6C1? Let us assume that the two guys are indeed in the same team. How many such teams can be formed? If these two guys are already in it, we have to select 1 guy from the remaining 6 guys to be part of the team. That can be done in 6C1 ways. This has to be 'removed' from the 8C3 ways of choosing 3 men out of 8 (8C3 is on an unrestricted basis). Hope this helps. Thanks.

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