$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples? We have a group of $16$ people, $8$ men, $8$ women.
We divide them to $8$ couples.
Let $X$ Be the number of couples make of $2$ men.
Calculate: $P(X = 3)$

I am not sure how to approach this.
I thought that this has an hypergeometric distribution.
Therefore i chose:
$$
N = 16, n = 16, K = 8, k = 6
$$
For:
$$
\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$
But its wont gonna work.
Another way i thought is to choose the couples like this:
$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{8}{1} \binom{7}{1} \binom{6}{2} \binom{4}{2}}{\frac{16!}{2^8}}
$$
The logic wass:

But again its not working.
I would like some help

Complicated problem.
Answer will be
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
For the denominator, not only do you have to consider how many ways there are of choosing the 1st couple, then the 2nd, ...
You then have to adjust for over-counting, re each grouping into 8 couples will be counted $8!$ ways.
Therefore,
$$D = \frac{\binom{16}{2}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{8!}.$$
For the numerator, first you have to create exactly 3 male couples.
So
$$N_1 = \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}}{3!}.$$
Then, you have to select among 8 women for the 1st odd man, and then among 7 women for the 2nd odd man.
$$N_2 = 8 \times 7.$$
Then, the 6 remaining women must pair up.
$$N_3 = \frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}.$$
Final answer is
$$\frac{N_1 \times N_2 \times N_3}{D}.$$
