The Gateway to Computer Science Excellence
+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. $\frac{8!}{(4!)^2}$
in Combinatory by Boss (16.8k points)
recategorized by | 17 views

1 Answer

0 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
by Junior (735 points)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,647 questions
56,492 answers
195,439 comments
100,703 users