The Problem
A tennis club has $4n$ members. To specify a doubles match, you choose two teams of two people. In how many ways can you arrange the members into doubles matches so that each player is in one doubles match? In how many ways can you do it if you also specify who serves first on each team?
Attempt at a Solution
So this problem is actually a follow up to this question. It comes from the same exact text as well. From the other problem we know that there are exactly $$\frac{(2n)!}{(n!)2^n}$$ ways to pair up $2n$ members for singles matches. I am tempted to simply double this number for $4n$ people, but that seems like it would be far to easy.
Questions and Confusion
While I am sure this problem abides by a similar logic to the one I linked, I am still struggling to understand a few key things.
- Do we consider the pairings $(p_1,p_2)$ and $(p_2, p_1)$ to be different? I feel like we aren't, but I cannot explain why…
- Here we are specifying doubles matches "so that each player is in one doubles match." Does this mean that each player can only be in one doubles match? Or does this simply mean to ensure that no player is left out of a match?
Thanks everyone!
Best Answer
Assume there are $n$ courts numbered from $1$ to $n$. You can send four people to each of these courts in $$A_n:={4n\choose 4}{4n-4\choose 4} \cdots{4\choose 4}={(4n)!\over (4!)^n}$$ ways. Since we are not interested in the numbers of the courts we have to divide by $n!$ in order to obtain the number of possible groupings of the $4n$ members into sets of four. Given such a set we can split it up into two teams in three ways.
The total number $D_n$ of arrangements of the $4n$ people into pairs of teams for doubles matches therefore is $$D_n={3^n\over n!}\>A_n={3^n(4n)!\over n! (4!)^n}={(4n)!\over n!\> 2^{3n}}\ .$$