The problems says
At the wedding of John and Mary there are $n$ men and $n$ women. In how many ways they can sit at a round table, so that no two men is next to each other and John and Mary sit together?
I have some doubts because of the answer to his problem, which doesn't match the one I got.
Let suppose that John and Mary are already seated and that people begin sitting from the side of Mary. We have two cases. One where the person next to Mary is a man and the other when it is a woman.
We can set the men en $n!/n = (n-1)!$ different ways.
The are $n-1$ gaps to put the women (since next to a man are Mary and John together. So there are $(n-1)!$ different ways to sit the women.
By the rule of product we have that there are $[(n-1)!]^2$ number of ways of sitting n men and n women beginning with a man next to Mary.
2nd Case (A woman sits next to Mary):
By the same reasoning as above we find that there are $[(n-1)!]^2$ number of ways of sitting them.
Finally by the rule of sum we have that there is a total of $2[(n-1)!]^2$ number of ways of sitting n men and n women having John and Mary always seated together.
What I got matches the answer of the textbook but I feel the reasoning I followed wasn't that right. The fact of having assumed that ''The are $n-1$ gaps to put the women..'' doesn't convince me. Certainly next to the first man is Mary (that is, the gap of that side is occupied) but nothing is telling that after putting the last men there will be no gap between him and John, so there may be anyway $n$ gaps to put the women