Combinatorics 8 men, 8 women, each choosing red or blue balls from bin My sister posed this question to me and I'm a little surprised to be struggling so much with it. 
8 men and 8 women choose balls from a bin that contains 8 blue balls and 8 red balls. After the selection is done, what is the probability of that 4 men have a red ball and 4 men have a blue ball?
I was able to brute force the problem for 2 men, 2 women, 2 red and 2 blue balls (answer is 2/3) and I wrote a little script to brute force the answer for 4 men, 4 women, 4 red and 4 blue balls (answer is 20736 / 40320).
I can't figure out how to generalize the count though. Any help would be much appreciated. 
 A: Imagine the $8$ balls to be distributed among the men are chosen at random,  with all choices equally likely. There are $\binom{16}{8}$ ways to choose these $8$ balls. And there are $\binom{8}{4}\binom{8}{4}$ ways for the men to choose $4$ red and $4$ blue. Divide.
A: You are distributing $8$ red balls among $8$ men and $8$ women (you can consider the blue ones as "not getting a red ball"). So the question is to check in how many ways you can select $4$ men and $4$ women, i e.:
$$\binom{8}{4} \cdot \binom{8}{4} = 
4900$$
A: Suppose that there are $2n$ balls of each color, $2n$ men, and $2n$ women. There are $\binom{4n}{2n}$ sets of balls that the $2n$ men might collectively pick, and $\binom{2n}n^2$ of them consist of $n$ balls of each color. Each set of $2n$ balls is equally likely to be chosen by the men, so the desired probability is
$$\frac{\binom{2n}n^2}{\binom{4n}{2n}}\;.$$
When $n=4$, this is
$$\frac{\binom84^2}{\binom{16}8}=\frac{70^2}{12870}=\frac{4900}{12870}=\frac{490}{1287}\;.$$
When $n=2$ it’s
$$\frac{\binom42^2}{\binom84}=\frac{36}{70}=\frac{18}{35}\;.$$
