Combinatorics: $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. > $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat? This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(< but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks! It is not necessary that women and men alternate, **AC** aB **cb**

Hint. By the inclusion-exclusion principle we have $$(2N)!-\binom{M}{1}\cdot\underbrace{2\binom{2N-1}{1}1!(2N-2)!}_{\text{at least one of the $M$ couples together}}+\binom{M}{2}\cdot\underbrace{2^2\binom{2N-2}{2}2!(2N-4)!}_{\text{at least two of the $M$ couples together}}-\dots$$ Can you take it from here?