Need a thorough explanation of this combination problem 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.
 A: 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.
A: 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.
A: 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 $\binom 6 3 \binom 5 2+\binom 52\binom 52$
A: I had the same problem of understanding this problem until I taught my own way of thinking I hope it helps.
Let's say that we split the number of men into -> 5 non-feuding men and 2 feuding men. Now we can form the committee with 2 women, 2 non-feuding men, and 1 feuding man or 2 women, 3 non-feuding men.
$$\binom{5}{2}\binom{5}{2}\binom{2}{1} + \binom{5}{2}\binom{5}{3} = 300$$
