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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?

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

Combinations can be 2m,4w
3m,3w
Since the least number of men is 2, and the least number of women is 3

For 2m,4w combinations..

Number of possible outcomes for having 4 women chosen from a group of 5 women = 5C4 = 5
Number of possible outcomes for having 2 men chosen from a group of 8 men = 8C2 = 28
But 1 pair refuse to serve together, we we're left with 28-1 = 27 pairs of men

So total number of possible outcomes for 2m4w combinations = 27*5 = 135

For 3m,3w combinations...
Number of possible outcomes for having 3 women chosen from a group of 5 women = 5C3 = 10
Number of possible outcomes for having 3 men chosen from a group of 8 men = 8C3 = 56
But 1 pair of men are fickled-minded, and so we have to remove 6C1 = 6 combinations, to give 56-6 = 50 pairs of men

Total number of possible outcomes for 3m3w combinations = 50*10 = 500

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...
_________________

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...

Oh oh, stop staring man. You seem to do this everyday evening in the beach.
I bet someone passed by when you skipped the last senetence.