How many ways are there ...

Question

There are $n$ women and $n$ men, where $n$ is divisible by 3. We want to divide them into groups of 3 members, but three men can't form a group, and three women can't form a group. Only groups of two women and a man, or of two men and a women are allowed. How many ways are there to do this?

Answer 1

If you just look at the women, you have a number of pairs and a number of singles. How many pairs and how many singles? Can you find the number of ways to group them that way? You have the same number of ways to group the men. Now you need to find the number of ways to pair the pairs of women with single men and vice versa.

Answer 2

We say a biased group is a pair of man and woman, along with a third person, written $((m,w),p)$. How many biased groups are there?

Let $3k=n.$ then choose $2k$ men. There are $\binom n {k}$ ways to do it. And choose $2k$ women: $\binom n {k}$ ways. Then for each man choose a woman, $(2k)!$ ways. Finally for each pair, choose a third person, $(2k)!$ ways. So there are $(\binom n k(2k)!)^2 = \left({n!\over k!}\right)^2$ biased groupings.

How many different biased groups are there for each group? Well there is one gender twice and one of those people can go in the first pair with the other as the third person. So each group corresponds to two biased groups. So $2k$ groups can be made out of $2k$ biased groups in $2^{2k}$ ways.

So the final answer is $$\frac{(n!)^2}{(k!)^24^k}$$

Answer 3

Let $n=3m$. Then there will be $m$ groups containing one man and two women, and there will be another $m$ groups containing one woman and two men. The $m$ single men, as well as the $m$ single women, can be chosen in ${3m\choose m}$ ways each. When these choices have been made each of the $m$ single men can choose two of the remaining $2m$ women in totally $$N_m={2m\choose2}{2m-2\choose2}\cdots{2\choose 2}={(2m)!\over 2^m}$$ ways, and similarly each of the $m$ single women can choose two of the remaining $2m$ men in $N_m$ ways. It follows that the total number $N$ of admissible partitions into $(2+1)$-groups is given by $$N=\left({3m \choose m}{(2m)!\over 2^m}\right)^2=\left({n!\over m!\cdot 2^m}\right)^2\ .$$