Number of pairwise disjoint subsets of size $k$ of a set of size $n$. I've been reading about combinatorics recently and have met the problem of choosing pairwise disjoint subsets of a set of size $n$.
I know the number of subsets of size $k$ of a set of size $n$ is given by the binomial coefficient ${n}\choose{k}$. So far I've read a bit about Stirling numbers of the second kind on Wikipedia and have seen a few answers to in-context problems on here but I haven't been able to find a good description of a general formula for the number of pairwise disjoint subsets and at least an overview of where it comes from.
I'm not sure if I should ask this as a separate question, but an example problem that I've been thinking about and haven't found a good answer for online (I've only found similar problems) is this:
Consider a group of $n$ tennis/badminton players. How many games of doubles can be picked from these $n$?
So far I can see that the number of ways of picking $4$ people for the game is ${n}\choose{4}$. I think the next thing to do is calculate the number of disjoint subsets of size $2$ of these $4$ (since one person can't be on both sides of the net and so in more than one of the subsets). Intuitively I think this should be $3$ but I can't see how to calculate it.
If it helps in giving answers I'm a first year mathematics undergraduate.
Thanks for the help!
 A: I'll provide my idea about why the answer should be $3$ to your example:
In your example, when you chosen two out of them, you implicitly chosen another two to be in the other team, so let's write down the formula for this part first
$(\textrm{choose two out of four into A team})\cdot(\textrm{implicitly choose another two into B team})$
$${4\choose2}\cdot{2\choose2}\ ,$$
But if I would understand your question right, e.g. say the four people are $a,b,c,d$, each of them, say $a$, will only care about whom will be his/her teammate, but he/she won't care about they will be assigned to which team(in this example only two, A and B), so team name doesn't matter.
Now the key is that


*

*all teams are of the same size, i.e. two-member team.

*we continuously choose players from the same group.


so the above formula will repeat. To correct this we divide it by $2!,$ since there are two teams, then your intuition is correct
$$(\textrm{ans. from the above})\cdot\frac{1}{2!}=3.$$
I'm not very familiar with Stirling number formula you may wait for others' answer.

Consider a group of n tennis/badminton players. How many games of doubles can be picked from these n?

If you just want to choose four out of $n$, then the answer could be
$${n\choose4}\cdot{4\choose2}{2\choose2}\cdot\frac{1}{2!}.$$
But if you would mean $n$ is a multiple of $4$ and want to consider all of them grouped into 2-2 pairs, then the answer is not this formula.
