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+1 vote

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. $\frac{8!}{(4!)^2}$
in Combinatory by
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1 Answer

+3 votes
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
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