Separate the numbers into pairs With how many ways can we separate the numbers $\{ 1,2,3, \dots, 2n\}$ into $n$ pairs, when:


*

*We don't care about the order of the pairs

*We care about the order of the pairs
$$$$
At the case when we don't care about the order of the pairs, is it $(\binom{2n}{2})^n$?
Is this correct?
How is it at the other case?
 A: We have say $10$ people, and want to divide them into teams of $2$. Arrange the people in order of student number, or height, or whatever.
The shortest person can choose her partner in $9$ ways. The first unchosen person then can choose her partner in $7$ ways. Then the first unchosen person can choose her partner in $5$ ways, and so on. 
More generally, the division into teams can be done in 
$$(2n-1)(2n-3)(2n-5)\cdots(3)(1)$$
ways. 
One can make this look nicer by multiplying and dividing by $(2n)(2n-2)(2n-4)\cdots (2)$. We get
$$\frac{(2n)!}{(2n)(2n-2)(2n-4)\cdots(2)}.$$
The bottom can be rewritten as $2^nn!$, so our number is
$$\frac{(2n!)}{2^n n!}.$$
For labelled teams, multiply by $n!$.
Another way: We first find the number of ways to divide the $2n$ people into labelled teams with leaders. Line up the people. This can be done in $(2n)!$ ways. Call people $1$ and $2$ Team $1$, with Leader $1$. Call people $3$ and $4$ Team $2$, with Leader $3$. In general people $2k-1$ and $2k$ are called Team $k$, with Leader $2k-1$. 
If $N$ is the number of leaderless labelled teams, then $N2^n=(2n)!$, since every division into leaderless labelled teams can be turned into a division into labelled teams with leader in $2^n$ ways. Thus $N=\frac{(2n)!}{2^n}$.
Finally, if $M$ is the number of unlabelled leaderless teams, then $Mn!=N$. Now we can find $M$.   
