$16$ 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

The number of ways to make $n$ couples from $2n$ people is
$$\frac{(2n)!}{n!\cdot 2^n}$$
This is derived by starting like you did, multiplying
$$\begin{align}
&\quad\binom{2n}{2} \cdot \binom{2n-2}{2} \cdot \binom{2n-4}{2} \cdots \binom{4}{2} \cdot \binom{2}{2} \\
\\
=& \quad\frac{2n(2n-1)}{2} \cdot \frac{(2n-2)(2n-3)}{2} \cdot \frac{(2n-4)(2n-5)}{2} \cdots   \frac{(2)(1)}{2} \\
\\
=&\quad \frac{(2n)!}{2^n}
\end{align}$$
However, the above includes an order for the couples chosen which we don't want, so the above has to be divided by $n!$
Now to count the number of ways of choosing exactly three male couples, we start by choosing the males to form the three couples, then apply the above formula once, then choose the two females for the two lone males, then apply the above formula a second time to pair up the remaining six females.
$$
\quad \binom{8}{6} \cdot \frac{6!}{3!\cdot 2^3} \cdot 8 \cdot 7 \cdot \frac{6!}{3!\cdot 2^3} = 352,800
$$
The number of ways to make eight couples with no restriction is
$$\frac{16!}{8!\cdot 2^8} = 2,027,025$$
Thus the required probability is
$$\frac{352,800}{2,027,025} = \boxed{\frac{224}{1287}}$$
