How many ways there are to divide $n$ couples ($2n$ people) to $n$ different rooms such that in each room there is exactly 2 people and couples can't be in the same room. (It is possible to have $2$ males or $2$ females inside a room).
It is probably inclusion-exclusion rule but I don't really understand how to use it.
On
As an alternative to Brian M. Scott's inclusion-exclusion answer, here is an approach that leads to a simple recursion formula.
Let $C_n$ count the number of ways that the $2n$ people comprising $n$ couples can be assigned to $n$ distinct rooms with $2$ people per room so that no couple is assigned to the same room. To help keep things straight, let's refer to the members of a couple as partners and reserve the word pair for the two people assigned to a room. So each person must be paired with someone but not with their partner.
Start with room 1: It can be assigned a pair in ${2n(2n-2)\over2}=2n(n-1)$ different ways. Having picked that pair, the remaining partners from Room 1's pair can either room together or apart. If they room together, they can do so in any of the $n-1$ remaining rooms and then the rest of the people from the remaining $n-2$ couples can be assigned to the remaining $n-2$ rooms, which can happen in $C_{n-2}$ ways, for a total of $(n-1)C_{n-2}$. But when we go to count the total number of ways in which they room apart, we might as well consider them to be a couple, in which case we have $n-1$ remaining couples to be assigned to $n-1$ remaining rooms, which can happen in $C_{n-1}$ ways. This gives the recursion
$$C_n=2n(n-1)((n-1)C_{n-2}+C_{n-1})$$
The recursion looks a little nicer if we "normalize" it by letting $C_n=n!D_n$, which makes sense since making the room distinct really just introduces an extra factor of $n!$. In this setting the recursion becomes
$$D_n=2(n-1)(D_{n-1}+D_{n-2})$$
Finally, to get things started, we need to note that $D_1=0$ (that is, if there's only one couple and one room, you can't split them up) and $D_2=2$ (or, if you prefer, $D_0=1$). This gives $0,2,8,60,\ldots$ for the $D_n$'s and $0,4,48,1440,\ldots$ for the $C_n$'s. The first of these is sequence A053871 in the OEIS, from which a good deal more can be learned. (Note, the sequence $0,4,48,1440,\ldots$ does not appear in the OEIS.)
For now I will assume that the rooms are interchangeable.
There are $\frac{(2n)!}{2^nn!}$ ways to pair up the $2n$ people; see this answer. Now number the couples from $1$ through $n$. For $k\in[m]$ let $A_k$ be the set of pairings that put couple $k$ in the same room. We want to avoid all of the pairings in $\bigcup_{k\in[n]}A_k$, so the answer to the question is
$$\frac{(2n)!}{2^nn!}-\left|\bigcup_{k\in[n]}A_k\right|\;.$$
We can use the inclusion-exclusion principle to compute $\left|\bigcup_{k\in[n]}A_k\right|$:
$$\begin{align*} \left|\bigcup_{k\in[n]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_k\right|\\ &=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\frac{(2(n-|I|))!}{2^{2(n-|I|)}(n-|I|)!}\\ &=\sum_{k=1}^n(-1)^{k+1}\binom{n}k\frac{(2(n-k))!}{2^{n-k}(n-k)!}\;, \end{align*}$$
since for each $k\in[n]$ there are $\binom{n}k$ subsets of $[n]$ of cardinality $k$. Thus,
$$\begin{align*} \frac{(2n)!}{2^nn!}-\left|\bigcup_{k\in[n]}A_k\right|&=\frac{(2n)!}{2^nn!}-\sum_{k=1}^n(-1)^{k+1}\binom{n}k\frac{(2(n-k))!}{2^{n-k}(n-k)!}\\ &=\sum_{k=0}^n(-1)^k\binom{n}k\frac{(2(n-k))!}{2^{n-k}(n-k)!}\\ &=\sum_{k=0}^n(-1)^{n-k}\binom{n}k\frac{(2k)!}{2^kk!}\;. \end{align*}$$
Since the rooms are in fact numbered, the actual answer desired is $n!$ times this.