**Hint:** We want to count the number of ways that we can choose $4$ people from the $12$ so that no two are married to each other. But then each person must belong to a different couple. So first, we choose the $4$ special couples from the $6$. Then for each couple, we choose $1$ of the $2$ partners. This yields: $$ \binom{6}{4}\binom{2}{1}^4 $$