Before we split into Case 1 and Case 2, $W_n$ is already paired with $M_j$.

You have no problem with Case 1, right?

Now consider Case 2.  $W_j$ is not paired with $M_n$.  We have the following people not paired yet: 

Women: $W_1, W_2, \ldots, W_{j-1}, W_j, W_{j+1}, \ldots, W_{n-1}$.  

Men: $M_1, M_2, \ldots, M_{j-1}, M_n, M_{j+1}, \ldots, M_{n-1}$.

We want to find the number of mismatches of this group, considering $W_j$ does not pair with $M_n$, it is equivalent to the original $n-1$ pairs and the newcomer $M_n$ is taking the place of $M_j$.  There are $a_{n-1}$ ways.