Don’t be so amazed, Melody. That happen every once in a while. LOL
My answer to Melody is:
Without the minus 1, the first set would be counted twice. Consider after each “main” permutation the couples are switched. This is a binary switch, and is demonstrable by using 5 coins. Start with all heads up then flip each in sequence on the tenth sequence you have returned to the beginning –all heads up. I’m still not sure if all the combined permutations are unique.
Considering Alan’s graphic I need to study this more.
The original permutation formula 4!*((5*2)-1)=216 should allow for all defined permutations, including reflections, where the only restriction is couples sit next to each other.
I need to study your graphic for awhile. One note is the M1 and F1 do not switch, where they would using the above formula. Even with the restriction of reflection only, the M1 and F1 should still switch—shouldn’t they?
This is curious. . . . We know the (n-1)! for the couples placements is correct, so question is the counting of the “sub- permutation” where the couples switch places. I may run computer simulations to examine the uniqueness of the outcomes. The formula I offered may only work in certain situations.
One thing about hyper-geometric distributions, they hyper-stretch one's brain.