How to find the number of ways to seat $n$ married couples around a table if men and women alternate

Question

Its my first time posting here, so if I am doing something wrong please inform me.

My question:

Find the number of ways to seat $n$ married couples around a table if men and women alternate.

I seen the solution where you seat $n$ men first and then the women. but I don't get why my solution is incorrect:

First we seat them in a row in the following way: we put a man in the first seat ($n$ ways), then a woman in the second ($n$ ways) and so on and we get $n!\times n!$.

Now there are $2n$ people on the row, to turn it into a table sitting we divide by $2n$ and we get $\dfrac{n!(n-1)!}{2}$ which is an incorrect answer.

I would like to understand where I made a mistake

The correct answer: $n!(n-1)!$

Edit: Just wanted to thank all of you! you really helped me there. If anybody have anything to add/more approaches to solving the problem - please do, I will gladly read it.

Answer 1

Note that with your approach you start by seating one of the $n$ men, rather than any of the $2n$ persons. That is, the $n!n!$ you initially come up with is the number of possible orderings for a straight table (or line-up) that start with a man. A such, when you take into account that the table is round, and that different rotations correspond to the same straight ordering, you should just divide by the number of men, rather than the number of total people.

Alternatively, you could first calculate the number of possible orderings for a straight line-up starting with any person. That means that you start with any of $2n$ choices ... but now that the gender is fixed, you have only $n$ choices for a person of the opposite sex for the second person, then $n-1$ choices for the third, $n-1$ for the fourth, etc. So, there are 2n(n−1)!n! possible straight orderings starting with man or woman. And now there are indeed $2n$ rotations per such ordering, so divide that by $2n$, and you get $(n−1)!n!$

Finally, let's do a concrete example to see what is happening. Suppose you have men $A$ and $B$ and women $C$ and $D$. Now, if you start the straight line-ups with a man, the possible straight orderings are $ACBD, ADBC, BCAD$, and $BDAC$ ( which is $4$ .. which is indeed $2!2!$). But how many circular orderings are there? There are $2$, so you need to divide by $2$ ... the number of men. Indeed, ACBD and BDAC are the same, and same for the other pair.

Alternatively, you could first consider all possible line-ups, starting with man or woman. So now you also have $CBDA, CADB, DACB$, and $DBAC$ as possible line-ups, for a total of $8$. And now you need to divide by $4$ (e.g. $ACBD, BDAC, CBDA$, and $DACB$ are all the same for a circular table)

Answer 2

@Bram already gave a nice answer concerning your misunderstanding. This answer only provides in another strategy to solve the problem.

For convenience assume that the men involved have distinct ages.

1) place the youngest man on a seat (it is inevitable that someone is the first right?).

2) place the remaining $n-1$ men on seats in such a way that no pair sits next to each other. This can be done on $(n-1)!$ ways.

3) place the women on the remaining seats. This can be done on $n!$ ways.

So there are $(n-1)!n!$ possibilities.