Three married couples have bought $6$ seats in a row for a concert. How many ways can they be seated if no man sits next to his wife.
I have worked through this problem and have got the correct answer. The problem I am having is, I can't, for the life of me, wrap my head around how to find the number of ways $1$ couple sits together.
$6!$ ways to order everyone $= 720$
All couples sit together: $3!(2!)^3=48$. Three different couples can be orientated $3!$ ways and each couple can be orientated $2!$ ways. Makes perfect sense to me.
Two couples sit together: $3(3)(2!)^4=144$. Pick which couple won't sit together, $3$. There are $3$ ways they cannot sit together and $2!$ ways to order them. $2!$ ways to order the remaining two couples and $2!$ ways to order which seat man or wife takes for both of them. Okay, great, I got that.
One couple sits together: I fumbled around with this until I came up with the $288$ I needed to get the correct answer: $3(2!)^2(4!)$. But, I have absolutely no idea how to make sense of it as I did above for the other possibilities. Pick a couple, $3$, order them $2!$. Pick a second couple $2$, then next to him can't be his wife, but it can be any of the other $3$. That gives you the $3(2!)^2$ and $3$. But we still need $4$ and $2$ of the $4!$. I can't reason out how we get to that conclusion.
Final answer: $720-48-144-288=240$. There are $240$ ways for no couple to sit together.
If someone could please break it down, Barney style, how one couple sitting together must be ordered it would be fantastic. I'm having a serious mental block.