How many ways are there to divide 2 groups separately into pairs, and then arrange the pairs? For example, there are $2$ groups of people, one made up of 8 females, the other $8$ males. So there will be $4$ pairs of females, and $4$ pairs of males. How many ways are there to arrange these $8$ pairs? (choice of pairs should also be considered)
Am I doing this correct? 
My solution is that there will be 
$$\left( \frac{\binom{8}{2\ 2\ 2\ 2}}{4!}\right)^2 \times 8!$$
total ways. Any thoughts would be appreciated.
 A: I will write down an explicit interpretation of the problem, since the intended meaning is not entirely clear. We want to arrange the $16$ people in a row, so that maximal contiguous groups of females have length $2$, $4$, $6$, or $8$, as do maximal contiguous groups of males. 
First we count the number of words over the alphabet $\{F, M\}$ in which maximal contiguous groups of  $F$ have length $2$, $4$, $6$, or $8$, and so do maximal contiguous groups of $M$. Let $\mathcal{F}$ denote a double $F$, and let $\mathcal{M}$ denote a double $M$. We want to count the words of length $8$ over the alphabet $\{\mathcal{F}, \mathcal{M}\}$. Such a word is determined once we decide where the $\mathcal{F}$ go. That can be done in $\binom{8}{4}$ ways. Multiply by $(8!)^2$ to take account of the $8!$ possible permutations of the females, and the $8!$ permutations of the males. 
Remark: Another possible interpretation is that (in the language of the answer) we have one of the sitting patterns $MMFFMMFF\dots FF$ or $FFMMFFMM\dots MM$. Then the number of arrangements is $2(8!)^2$. I have not thought of a reasonable interpretation that yields your answer. If you edit your post to give your interpretation of the problem, I can check whether your expression is a correct count for that interpretation.
