How to divide 2n members of the club into disjoint teams of 2 members each, when teams not labelled?

Question

I am doing this and getting $\sum_{i=0}^{2n-2} \frac{(2n-i)!}{2!(2n-i-2)!}$ but the answer given is $\frac{(2n)!}{2^n n!}$ I even tried to get this by simplifying my result, but not getting the same.

Answer 1

Their answer: imagine lining up everyone in a row and then pairing up adjacent people. There are $(2n)!$ ways to line everyone up, but different line-ups can produce the same teams. To account for this overcounting, divide by $n!$ for the number of ways to arrange the $n$ teams in a row, and divide by $2$ for each pair to account for swapping the order of the members of each pair.

In your answer, you probably meant $\prod_{i=0}^{2n-2} \binom{2n-i}{2} = \prod_{i=0}^{2n-2}\frac{(2n-i)(2n-i-1)}{2} = \frac{(2n)!}{2^n}$, which comes from choosing the first pair, choosing the second pair, and so on. But here, you are accounting for the order that the teams are chosen. (Recall the problem states the teams are not labeled.) So, you should divide by $n!$. This agrees with their answer.

Answer 2

There is in fact a direct formula for this:
Formula: The number of ways in which $mn$ different items can be divided into $m$ groups each consisting of $n$ objects, with order not important is given by: $$ \frac{(mn)!}{(n!)^{m}\times m!}. $$ In your problem, with $m=n$ and $n=2$ we have $\frac{(2n)!}{(2)!^{n}\times n!}$ which is the answer!

Answer 3

Your analysis leads to the $\frac{2n!}{2^n}$ part of the answer, but overlooks the overcount due to the order of the groups.

As a direct method, consider simply starting with all orderings of the $2n$ members ($2n!$ options) and making groups by first $2$, next $2$ etc. This overcounts in two ways

• the order of the members in the pairing
• the order of the pairs

Dividing out these overcounts gives the result $\dfrac{2n!}{2^nn!}$

Answer 4

How did you proceeded??

I am getting the same answer as given.

My way is

For choosing 2 members out of 2n,, the ways are $2n\choose 2$

Again choosing 2 members out of 2n-2 ,the ways are $2n-2\choose 2$

Proceeding this way you will get

$2n\choose 2$.$2n-2\choose 2$.$2n-4\choose 2$...$2n-(2n-4)\choose 2$.$2n-(2n-2)\choose 2$.$2n-2n\choose 2$

On solving you will get $\frac {(2n)!}{2^nn!}$

That is the solution

Answer 5

Look at the problem of arranging letters in the word MAMA. It is $\binom{4}{2}$, because it enumerates all ways to get 2 disjoint subsets froma set of 4 numbers and compare to the case when Ms and As are of different colors: you need to mulfiply it by 2!. The first case is your situation.