+1 vote
17 views

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

1. every team has exactly $2$ members, and
2. 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
1. $\frac{8!}{2^4}$
2. $\frac{8!}{2^4(4!)}$
3. $\frac{8!}{4!}$
4. $\frac{8!}{(4!)^2}$

recategorized | 17 views

Total 8 members are there so 8C2*6C2*4C2*2C2 are total ways but here order of teams does not matter so divide by 4! there 8!/2^4*4! so option B is the answer
by Junior (735 points)

+1 vote