Sitting $n$ men and $n$ woman around a table having the wedding couple seated together 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.
My attempt:
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.
1st Case:
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
 A: First of all the question is a little bit ambigous. We don't know whether the table is a circle of a row, but from your reasoning it can be concluded that it's a circle, although not for certain. Also are there $n$ men and $n$ women total or $n$ men and $n$ women plus John and Mary? Considering the textbook answer I would go with the first option. Anyway from now on I will accept the mentioned assumptions as true.
Note that there have to be an empty space between each men. Or to make it more simple if there are $2n$ numbered seats man can sit in the even or odd seats. So we have to put $n$ men in $n$ seats around a round table. The number of possibilities is $(n-1)!$ 
Now as Mary has to sit next to John we have two options for her, left of John or right of him. Next the remaining $n-1$ women can be seated in $n-1$ seats in $(n-1)!$. Eventually as the events are not related we have that the total number of combinations is $2\left((n-1)!\right)^2$
I guess this explanation will fill the "holes" in your solution.
A: The rule that men and women alternate should also apply to John and Mary. Hence only your 1st case applies, i.e. $[(n-1)!]^2$, and not your 2nd case.
However, you should also consider the "mirror configuration" of the 1st case, i.e. where John is seated on Mary's left/right. 
Hence the number of configurations should be multiplied by $2$, giving $2[(n-1)!]^2$.

NB - this is not the same as beads on a ring where you can flip over the ring  and consider the configuration to be the same - here you can't flip over the dinner table!
