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From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed?

This one's easy. There's two experiments (ex 1 = committees of men)(ex 2 = committees of women) so it's just $5 \choose 2$$7 \choose 3$.

But the next question is

What if 2 of the men are feuding and refuse to serve on the committee together?

I don't understand this question at all.

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  • $\begingroup$ Can you be more specific? What don't you understand about it? $\endgroup$ – Nick Peterson Sep 7 '13 at 17:53
  • $\begingroup$ I don't understand the intuition behind it, or how to start. $\endgroup$ – Don Larynx Sep 7 '13 at 17:58
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As counterpoint to Ross's forward solution, here is a backward solution. You've already counted all the committees, now let's subtract the "bad" ones, where two feuding men are serving together. You also need two women and a non-feuding man, so there are $${5\choose 2}{5\choose 1}$$ bad committees, which you can subtract from $${5\choose 2}{7\choose 3}$$ to find your answer.

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Hint: That means you have to calculate how many groups of three men do not include both of the pair. I would say you now have three groups of people-women, feuding men, and non-feuding men. You need two women, one feuding man, and two non-feuding men or two women and three non-feuding men. You calculate each of these in the way you did for women and men.

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  • $\begingroup$ How would you put this in an equation? $\endgroup$ – Don Larynx Sep 29 '13 at 19:42
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    $\begingroup$ @DonLarynx: It would be ${5 \choose 2}({2 \choose 1}{5 \choose 2}+{5 \choose 3})$ $\endgroup$ – Ross Millikan Sep 29 '13 at 21:16
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Here is another way of counting it. Treat the feuding men as a single man, and calculate $\binom 6 3 \binom 5 2$ as the base number of possibilities.

But each of the combinations with a feuding man in should have been counted twice (it could have been either man). That is $\binom 52\binom 52$ to add - because we have to find two non-feuding men and two women.

So the total is $\binom 6 3 \binom 5 2+\binom 52\binom 52$

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