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

Question

I thought about this:

(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 $(2n)^{2n}$. — Arthur, May 14 '13 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... — Jakub Konieczny, May 14 '13 at 19:39
One minor thing: last sum is either from $k = 1$, or skip the term $2n!$ in front. — Jakub Konieczny, May 14 '13 at 20:07
You actually can just do the sum from $0$ and omit the $(2n)!$ — NightRa, Sep 03 '13 at 17:06
Doesn't $(2n-1)!$ include a permutation where a husband is next to his wife? I do understand that there are $(2n-1)$ ways to arrange the remaing people after choosing a couple $\binom n1$. I don't see why the $(2n-1)!$ won't include a couple who are adjacent. — Faizal Ismaeel, Nov 22 '14 at 01:42
You're right. If you want more problems with answers check Counting: The art of enumerative combinatorics by G. Martin. — user2820579, Mar 16 '16 at 19:05
Possible duplicate of Arranging couples (husband and wife) on a bench so no wife would sit next to her husband? — inavda, Aug 18 '18 at 16:09
@inavda The question you linked to is more recent. Wouldn't it make more sense to close in the other direction? — Arnaud D., Aug 19 '18 at 12:39
@ArnaudD. The other question has a more complete answer so I elected to close this one. — inavda, Aug 20 '18 at 15:46
Related: https://math.stackexchange.com/questions/465318/showing-probability-no-husband-next-to-wife-converges-to-e-1 — Henry, Jul 08 '21 at 00:50

score 0 · Accepted Answer · answered May 14 '13 at 20:06

As far as I am able to tell, the answer is correct.

After some unsuccessful attempts at simplifying the resulting expression, I went ahead and asked Wolfram for the answer. He came up with the formula:

$$ 4^n \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}} \operatorname{Hypergeometric1F1}(-n,-2n,-2)$$

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".

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

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