The number of ways in which 4 married couples can be seated such that 4 persons are on left side and 4 are on right side of a long table and no wife is in front of her husband and same gender persons are not sitting opposite to each other

Here's what I know. We can select 4 people on the left. It can be all men,3men 1women,2men 2 women, 1men 3 women and all women. And for each person we have to select a person for opposite gender except his wife.

How will I now pair them up with their opposite gender but not their spouses

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    $\begingroup$ Count the number of ways of seating the men, then multiply by the number of derangements of their wives. $\endgroup$ – N. F. Taussig Aug 20 at 10:26
  • $\begingroup$ A similar approach to the one @N.F.Taussig proposed, but looks easier at least to me, solve the special case where all men sit right, and then count the swaps to revert back to the general case. $\endgroup$ – cgss Aug 20 at 10:37

Following @cgss's comment, there are $4!=24$ ways to seat the men on the right side.

For each such arrangement of men on the right, the number of ways to seat the women is the number of derangements for a set of size $4$, or equivalently, the number of permutations of {$1,2,3,4$} which have no fixed points. These include the $6$ many $4$-cycles and the $3$ many products of two $2$-cycles, for a total of $9$ ways to seat the women opposite a fixed order of the men on the right.

(If you haven't seen derangements or permutation groups, you could list all $9$ by writing them down). The $15$ "bad" arrangements include $8$ with exactly one couple seated opposite each other, $6$ with two couples each opposite each other, and $1$ with every husband and wife opposite each other)

So there are $24*9$ ways to seat them with all the the men on the right. Now for each such way, you can flip (or not flip) any opposite pair and still be admissible, so there are $2^4=16$ ways to do it for each of the $24*9$ "all men on the right" arrangements.

So you get $24*9*16$ arrangements altogether.

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