# In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?

In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?

(Total amount of ways to sit 2n people in 2n sits)-(Using inclusion and exclusion to find that at least 1 wife sits next to her husband) And i get:

Let $A_1$ be the attribute where at least 1 wife sits with her husband, Then we "merge" up the husband and wife into a one person. We have $\binom n1$ ways to choose $1$ couple out of $n$ couples, And we are left with $2n-1$ to place $2n-1$ 'people' so we get $(2n-1)!$ and so on, And on a general note:

$(2n)!-(2\binom n1(2n-1)!-2^2\binom n2(2n-2)!+...2^k(-1)^k\binom nk(2n-k)!)$ And a bit simplified:

$(2n)!-(\sum_{k=1}^n2^k(-1)^k\binom nk(2n-k)!$)

Since i don't have answers to this question i wanna know if i did something wrong? Did i even look at the question right?

• $2n$ people can sit in $2n$ seats in $(2n)!$ ways, not $(2n)^{2n}$. May 14, 2013 at 19:37
• Don't you want to have factors like $2^k{n \choose k}$ in place of ${n \choose k}$ ? You can sit husband and wife next to each other in two ways... May 14, 2013 at 19:39
• One minor thing: last sum is either from $k = 1$, or skip the term $2n!$ in front. May 14, 2013 at 20:07
• You actually can just do the sum from $0$ and omit the $(2n)!$ Sep 3, 2013 at 17:06
• Aug 18, 2018 at 16:09

$$4^n \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}} \operatorname{Hypergeometric1F1}(-n,-2n,-2)$$
The weird looking term $\frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}}$ is just $(2n-1)!!/2^n$, so this can also be written as:
$$2^n (2n-1)!! \cdot \operatorname{Hypergeometri1cF1}(-n,-2n,-2)$$
Unfortunately, it seems that $\operatorname{Hypergeometri1cF1}(-n,-2n,-2)$ is not anything nicer. Of course, I only have heuristic justification of the type "if it was something nicer, it wouldn't be called with such scary name, and Wolfram would be able to simplify it".