Find the number of ways $z_n$ of seating $n$ couples around a rectangular table such that no one is allowed to sit next to his or her partner Find the number of ways $z_n$ of seating $n$ couples around a rectangular table such that no one is allowed to sit next to his or her partner.figure $(\text{I})$.
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$$\text{Figure (I)}$$
First we should find the number of ways that $2n$ people can sit around the table,we choose $n$ of $2n$ people to sit on one of the sides of the table in $\binom{2n}{n}$ ways,besides for the people sitting on each sides of the table there are $n!$ permutations,and so by the multiplicative law:$$\binom{2n}{n}\left(n!\right)^{2}=\left(2n\right)!$$
Denote by $w_k$ the number of seatings under which some specified set of $k$ couples (and possibly some other couples) end up sitting next to their partner:
$$z_n=\left|\bigcap_{i=1}^{n}\overline{A_i}\right|=\left(2n\right)!-\left|\bigcup_{i=1}^{n}A_i\right|=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}w_{k}$$
Now it's left to determine a formula for $w_k$:
This is where I cannot continue,I thought that the formula maybe is :
$$w_k=\binom{2n}{2k}k!\cdot2^{k}\left(2n-2k\right)!$$
(Decide where the k couples go, and which couple goes where, and which partner takes which seat, and where the $2n-2k$ individuals go.) 
However after some thought,I figured out that this is not true,since it may happen that one of the husbands/wives be left unpaired,so what is the strategy to solve the problem?
 A: The number of ways of forming $k$ non-overlapping pairs of adjacent seats is
$$
\sum_{r=0}^k\binom{n-r}{r}\binom{n-(k-r)}{k-r}.
$$
The two binomial coefficients come from the answer to this question applied with parameters $n-1$ and $r$ for the seats on the front side of the table and parameters $n-1$ and $k-r$ for the seats on the back side of the table. The reason for $n-1$ is that the leftmost chair in a pair can't be the rightmost seat in the row.
Using this result and the principle of inclusion-exclusion, the number of arrangements is
$$
\sum_{k=0}^n(-1)^k\frac{n!}{(n-k)!}2^k(2n-2k)!\sum_{r=0}^k\binom{n-r}{r}\binom{n-(k-r)}{k-r}.
$$
In this expression, $\frac{n!}{(n-k)!}$ is the number of ways of assigning couples to the chosen pairs of seats, $2^k$ is the number of ways of assigning members of the couples to seats, and $(2n-2k)!$ is the number of ways of assigning the remaining individuals to seats.
For $n=0$, $1$, $2$, $3, \ldots$ the values of this expression are
$$
1,\ 2,\ 16,\ 336,\ 16512,\ 1428480,\ \ldots
$$
