In how many ways, order matters, can 14 married couples be seated in chairs consecutively 1 to 28 at a round table so that one man is always between two women and no man ever sits next to his wife?
Intriguing isn't it.
My way of looking at the problem
My attempt at a calcultation
Our third choice cannot be equal to our first or second, therefore n-2, likewise our fourth is n-2 since it cannot be equal to the second or third. Seeing a pattern yet?
I believe that we arrive at n(n-1)(n-2)(n-2)(n-3)(n-3).... or n(n-1)((n-2)!)^2. For the given parameters this is about 1.38*10^24, so there are quite a few ways to sit the party it would seem.
Let me know where I messed up!