Counting ways of seating couples, keeping men apart, women apart, and couples apart 
There are $n$ man-woman couples sitting around a table, with seat number $1,2,\dots,2n$. How many ways of sitting that keeps men apart, women apart, and simultaneously couples apart?

I intend to tackle the problem invoking the Inclusion-Exclusion Principle.  
Set ground set 
$$X= \{ \text{sittings that keep men apart and women apart}\}$$
Label the couples with number $1$, $2$, $\dots$, $n$. Define 
$$A_i = \{ \text{sittings that keeps men apart and women apart, but keep the $i$-th couple together} \}$$ 
for each $i\in[n]$. 
Then the number of sitting we set out to find is equal to:
$$\biggl|X-\bigcup_{i=1}^nA_i\biggr|=\biggl|\bigcap_{i=1}^nA_i^c\biggr|$$
The only problem left is how to determine size of $A_I$, where $I$ is a subset of $[n]$. But I have no idea how to enumerate this. Anybody got any ideas?
 A: This is only a partial answer, which only reduces this enumeration to another one. 

For a round table with $n$ seats, call an $(n,k)$-template a combination of non-intersecting pairs of adjacent seats; I hope this notion will get clearer with the following discussion. Let $t(n,k)$ be the number of $(n,k)$-templates; here is a table of some values: 
$$
\begin{array}{c|ccccc}
&0 & 1& 2& 3& 4 \\\hline 
1& 1\\
2& 1& 1\\
3& 1& 3\\
4& 1& 4& 2\\
5& 1& 5& 5\\
6& 1& 6& 9& 2\\
7& 1& 7& 14& 7\\
8& 1& 8& 20& 16& 2\\
9& 1& 9& 27& 30& 9
\end{array}
$$
Here are some interesting things about it. 


*

*In most cases, $t(n,k) = t(n-1,k) + t(n-2,k-1)$

*The row sum (the number of $n$-templates) is $n$th Lucas number (except $n=2$ for some obvious reasons).

Assume that the sex of one person at the table is fixed (to avoid multiplying by $2$ everywhere). 
Then $|X| = (n!)^2$. Define
$$
B_i = \{\text{the persons sitting on $i$th and $(i+1)$th seats are a pair}\}.
$$
(I identify $2n+1$ and $1$.)
Clearly, $$
\bigl|B_{i_1}\cap B_{i_2}\cap \dots\cap B_{i_k}\bigr| = n\cdot (n-1)\cdots (n-k+1)((n-k)!)^2 = n! (n-k)!
$$ 
(choose couples for the seat pairs, then sit the remaining people)
provided that the pairs don't intersect, i.e. $\{i_j,i_j+1\}\cap \{i_l,i_l+1\} = \varnothing$, $j\neq l$; otherwise the intersection is empty. But the number of such non-intersecting pairs is exactly $t(2n,k)$.
Therefore, 
$$
\biggl|\bigcup_{i=1}^n B_i^c\biggr| = |X| - \sum_{k=1}^n(-1)^{k-1} \sum_{1\le i_1<\dots<i_k\le n} \bigl|B_{i_1}\cap B_{i_2}\cap \dots\cap B_{i_k}\bigr|=\\
= \sum_{k=0}^n (-1)^k t(2n,k) n! (n-k)! = n! \sum_{k=0}^n(-1)^k t(2n,k) (n-k)!
$$
Well, nothing very pleasant, but I find the question of enumerating $t(n,k)$ very interesting by itself (It must be related to certain statistical physics problems). 

Following @awkward's comment, I found here that $t(n,k) = \frac{n}{n-k}{n-k\choose k}$ (however, this is not absolutely correct, as $t(2,1) = 1$, not $2$; otherwise seems good). The link also describes the solution better than I wrote.
