In how many ways $n$ couples can seat around a table with men and women alternating? In how many ways  $n$ couples (heterosexuals) can seat around a table with men and women alternating?
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If the number of tables of both sides is an odd number then we need to select $\binom{n}{\lfloor n/2\rfloor}$ men and $\binom{n}{\lfloor n/2\rfloor+1}$ women to sit on one of the sides,,besides for the people sitting on each sides of the table there are $(\lfloor n/2\rfloor+1)!(\lfloor n/2\rfloor)!$ permutations,and so by the multiplicative law:
$$\binom{n}{\lfloor n/2\rfloor}\binom{n}{\lfloor n/2\rfloor+1}(\lfloor n/2\rfloor+1)!^2(\lfloor n/2\rfloor)!^2=n!^2$$
If the number of tables of both sides is an even number then we need to select $\binom{n}{n/2}$ men and $\binom{n}{n/2}$ women to sit on one of the sides,moreover woman and men can change their place in $2$ ways,,besides for the people sitting on each sides of the table there are $(n/2)!(n/2)!$ permutations,and so by the multiplicative law:
$$4\binom{n}{n/2}^2(n/2)!^4=4(n!)^2$$

However I'm not sure about the answer,can someone please check that?
 A: When counting "no. of ways" it is important to determine the equivalence classes, i.e. what counts as "different"?  In this case, if you have a particular seating, and then you rotate the whole table $180°$, is that the same seating or a different seating?  Your answers seem confused on this point.
First, assume the rotated table counts as a different "way".  Then:


*

*Your answer for even $n$ is correct: $4 \times n!^2$ 

*But your answer for odd $n$ is incorrect: You forgot a factor of $2\times$ because you have not specified whether the more-men or the more-women side is at the north side of the table.  So the odd $n$ answer should be: $2 \times n!^2$.
Alternately, assume the rotated table counts as the same "way".  Then:


*

*Your answer for odd $n$ is correct: $n!^2$, because you have specified what happens on the more-women side and what happens on the more-men side, and you don't need to care which side is north (due to rotation equivalence).

*But now your answer for even $n$ has double-counted: you have counted both a seating and its rotation, so you need to divide by two: $4 \times n!^2 / 2 = 2\times n!^2$.
In short, one of your answers is right while the other is wrong, but which is right/wrong depends on interpretation of what counts as "different" seatings.
BTW, there is truly a factor of $2 \times$ between the even answer and the odd answer, due to the extra degree of freedom of seating men opposite men (across the table) or men opposite women in the even case.  There is just not a factor of $4\times$ difference.

Incidentally, it is often a good idea to check your answers against small examples, e.g. $n=1, 2$.  Once you do that you will see right away you needed to first answer the question of whether rotation counts as different.  E.g. suppose we say rotations are different:


*

*For $n=1$, Peter can sit north, or south, so there are $2 \times 1!^2 = 2$ ways.

*For $n=2$, Peter can sit in any of four seats, then George has two choices, then we can pick between Jane and Mary to be next to Peter, so there are $4 \times 2 \times 2 = 16 = 4 \times 2!^2$ ways.
