In a room there are $8$ men, numbered $1,2, \dots ,8$. These men have to be divided into $4$ teams in such a way that
- every team has exactly $2$ members, and
- there are no common members between any two teams. For example, $\{(1,2),(3,4),(5,6),(7,8)\} \{(1,5),(2,7),(3,8),(4,6)\}$ are two such $4$-team combinations. The total number of such $4$-team combinations is
- $\frac{8!}{2^4}$
- $\frac{8!}{2^4(4!)}$
- $\frac{8!}{4!}$
- $\frac{8!}{(4!)^2}$