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