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It is required to divide the $2n$ members of a club into $n$ disjoint teams of 2 members each. The teams are not labelled. The number of ways in which this can be done is:

1. $\frac{\left ( 2n \right )!}{2^{n}}$
2. $\frac{\left ( 2n \right )!}{n!}$
3. $\frac{\left ( 2n \right )!}{2^n . n!}$
4. $\frac{n!}{2}$
5. None of the above.
edited | 604 views
+1
Choose n out of 2n members = 2nCn ways.
Remaining n members can be divided into n teams in n! ways.
So total = (2nCn)n! ways
But in each team, (m1,m2) = (m2,m1), so we have to divide by 2 'n' times i.e 2^n.

This way of reasoning is correct?

${2n}$ member to be n teams with $2$ member each and teams are unordered so we can exchange $n$ team member among them.

=$\frac{(2n)!}{\underbrace{2!.2!.2! \dots 2!}_{n \text{ times }} \times n!}$
=$\frac{(2n)!}{2^n \times n!}$

Option c.
edited
+1

i didnt understand the "/n!" part

conflict with gateoverflow.in/25431

+7
because the teams are not labeled here. So, it is like we have n teams but the cases where the same set of players going to two different teams are the same. i.e., it is only the players which distinguishes the team (not the team name). So, this is equivalent to considering all the possible permutations on n teams as 1.

An almost same question with different answer -> this is what they ask for GATE/TIFR.
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If we number the players from 1,2,3,........2n.

In terms of this what does this( The teams are not labelled. ) mean?

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@sumit , only the players which distinguishes the team, not the team name so (1,2 )is a team , (3,4) another team wthut any odering
+3

Behind The Scenes

Let n=3 Members={a1,a2,a3,b1,b2,b3}

Total 6 players.

If the Teams were labeled Then things we go like this

 A B C (a1,a2) (b1,b2) (a3,b3) (a1,a2) (a3,b3) (b1,b2) ...... ...... .......

What above table says that in 3 teams A, B, C we put our team players in those pairs i.e. (a1,a2)(b1,b2)(a3,b3).
Then among A, B, C can arrange these pairs in 3!=6 ways. Correct? that is why 6 rows in the table.

But as soon as I'll remove the label A, B, C from them, all these 6 arrangements become 1. i.e. means we need to divide by 6 the total no. ways to arrange 6 members into 3 teams.

Similarly, when we remove labels from n-teams in which 2n members were arranged, we need to divide by n!.

+1
or in other terms

$\binom{2n}{2} \binom{2n-2}{2}\binom{2n-4}{2}.......\binom{2n-2(n-1)}{2}$