For the following problem, I feel my reasoning is something wrong, so I would like if it is in the right direction or if it needs to be rephrased/corrected.
The problem reads:
How many ways are there to sit $n$ wife-husband couples in a row with no wife anywhere to the right of her husband if men and women alternate.
The first step is to arrange either the men or the women in a row: $n!$.
Now I choose an arrangement, perhaps the easiest one and try to accomodate a women in each space (2 dots indicate an available space):
$$ ..m_1..m_2.. (etc)..m_n $$
Now, to the left of $m_1$ we can arrange any women. Now, suppose 2 cases:
(a) At first place, $w_1$ is not placed. Thus, to the right of $m_1$ there are $n-2$ options, to the right of $m_2$ there should be $n-2$ options, because here you can fit $w_1$.For this arrangement, at the end of the day there should be $(n-2)(n-1)!$.
(b) We place $w_1$ at the first place. to the left of $m_1$ there are $(n-1)$ options, to the right of $m_2$ $n-2$ options. Here we have $(n-1)!$ options to place the women.
(c) Here we choose to start the line with the men:
$$ m_1..m_2.. (etc)..m_n.. $$
To the right of $m_1$ there are $(n-1)$ options, then $(n-2)$, etc. Total here: $(n-1)!$
Answer:
$$ n! * ((n-2)*(n-1)! + (n-1)! + (n-1)!) = n!^2 $$
Last question: How this problem is different from the problem:
How many ways are there to seat $n$ wife-husband couples in a row with no wife next to her husband on his right if men and women alternate?
To me, they both look the same.