$16$ 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
 A: Choose 6 men to be paired with one another. This can be done in $\binom 86=28$ ways. Then pair them up. This can be done in $5!!=15$ ways. Then take the remaining two men, and pair them up with two women. This can be done in $_8P_2=56$ ways. Finally, pair up the remaining six women. This can be done in $5!!=15$ ways.
In total, there are
$$
28\cdot 15\cdot56\cdot 15=352\,800
$$
ways to make exactly three male couples. Finally, to get the probability, divide by the $15!!=2\,027\,025$ ways to make 8 couples.
A: 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
$$\frac{N_1 \times N_2 \times N_3}{D}.$$
A: 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
$$\frac{352,800}{2,027,025} = \boxed{\frac{224}{1287}}$$
