dividing a group 0f 10 men and women into couples of same sex 
There're 10 men and 10 women. We need to divide them into couples of same sex. 

The answer is: $$\left( \frac{10!}{5!*2^5} \right)^2$$ I don't understand why. 
I know that we can divide into couples where there's exactly 1 man and 1 woman in each couple in $10!$ ways (1). Now I would need to find the total possible number of couples regardless of sex $\left( \frac{20!}{10!*2^{10}} \right)$ (2). Then (2) - (1) should give the answer. But that doesn't give the same result.
 A: An easier way to get the number of possibilities : If we fix a man, we have $9$ choices for the first couple. If we fix a man again, we have $7$ choices, then $5$, then $3$. 
In total, we have $9\cdot 7\cdot 5\cdot 3$ possibilities. 
Same with the women, so finally we have $(9\cdot 7\cdot 5\cdot 3)^2$ possibilities.
A: Although you didn't write the formula for the answer, it's not too difficult to guess it. For just one group of $2x$ persons, we get:
$$c(x) = \frac{2x!}{x! \cdot 2^x}$$
For the example with $10$ men and $10$ women, this would be $(c(5))^2$
And that gives you the result for all combinations as well, as you wrote $\frac{20!}{10! \cdot 2^{10}}$. But don't subtract anything from that - there is no reason to.
A: $10!$ is the number of different ways to arrange the women.
Each arrangement determines the coupling (1st with 2nd, 3rd with 4th, etc).
Divide by $5!$ in order to eliminate different arrangements of each coupling.
Divide by $2^5$ is in order to eliminate different arrangements of each couple.
Raise the result to the power of $2$ in order to calculate the same for the men.
A: You are right about the total possible number of couples regardless of sex.  The number $\frac{20!}{10!*2^{10}}$ includes every possibility: out of the 10 couples, they can be of the same gender, or some of them can be of 1 man and 1 woman.  The number $10!$ represents the case in which none of the couples are of the same sex, so you can subtract $10!$ from that to get the answer.
We first divide them into two groups of 10 persons by sex.  In each of these two groups,
we have $\binom{10}{2}$ ways to choose the first couple while leaving $8 \times 2$ persons behind, then we have $\binom{8}{2}$ ways to pick another couple while leaving $6 \times 2$ persons behind.  We continue this process until everyone has its partner.  Therefore, we have
$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$, but we need to divide this by $5!$ since we ask for the number of ways of dividing them into couples.  This gives the desired result for the group of 10 persons.  Squaring this will give the required answer.
\begin{align}
\frac{1}{5!}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} &= \frac{1}{5!} \frac{10\times9}{2}\,\frac{8\times7}{2}\,\dots\frac{2\times1}{2} \\
&= \frac{10!}{5!*2^{5}}
\end{align}
