4 married couples in a circular arrangement where no unmarried couples are adjacent 
Given 4 married couples. They will be arranged around a circular table where no unmarried couples are adjacent.

Attempt

*

*Arrange the 4 men around the table first. There are $(4-1)!=6$ circular permutations.

*Choose two adjacent men (among 4 possible ways) and insert the remaining 4 women such that the women at the end matches her husband. The two women in the middle can be swapped to produce another unique circular permutation. In this step there are $4\times 2=8$ ways.

Thus in total there is $6\times 8=48$ circular permutations.
Question
Is my calculation correct?
 A: Suppose that the $4$ men have been seated. The restriction on unmarried couples means that if there are any women between two adjacent men, there must be at least two: the wives of the two men. That is, one can have a sequence $M_1W_1W_2M_2$, and either or both of the other two women can sit between $W_1$ and $W_2$. However, suppose that only $W_3$ does so, making the sequence $M_1W_1W_3W_2M_2$: then $W_4$ will be forced to sit next to a man who is not her husband. Thus, if there are any women between $M_1$ and $M_2$, they must be either $W_1$ and $W_2$ or all four women, and the possible orders are $M_1W_1W_2M_2$ and $M_1W_1W_kW_\ell W_2M_2$, where $k$ and $\ell$ are $3$ and $4$ in either order.
In the first case the whole arrangement must take the form $M_1W_1W_2M_2M_kW_kW_\ell M_\ell$, where $\{k,\ell\}=\{3,4\}$. In the second it must be $M_1W_1W_kW_\ell W_2M_2M_mM_n$, where $\{k,\ell\}=\{m,n\}=\{3,4\}$. You counted the arrangements in the second case but not those in the first case. In the first case there are again $6$ ways to seat the men. There are $2$ ways to choose which pairs of men will have women sitting between them, and the seating of the women is then forced, so there are $12$ possible arrangements of this type, for a total of $60$ altogether.
