Consider a game, with $n$ players, $n$ even. Each match involves exactly 2 players. I want to compute the total number of combinations you can have at first round. For example, for $n=2$, just 1 combination. For $n=4$, you can have $3$:
1) player1 vs player2, player3 vs player4
2) p1 vs p3, p2 vs p4
3) p1 vs p4, p2 vs p3
For $n=6$ you can have 15 combinations:
1) (1,2), (3,4), (5,6)
2) (1,2), (3,5), (4,6)
3) (1,2), (3,6), (4,5)
4) (1,3), (2,4), (5,6)
5) (1,3), (2,5), (4,6)
6) (1,3), (2,6), (4,5)
7) (1,3), (2,3), (5,6)
8) (1,4), (2,5), (3,6)
9) (1,4), (2,6), (3,5)
10)(1,5), (2,3), (4,6)
11)(1,5), (2,4), (3,6)
12)(1,5), (2,6), (3,4)
13)(1,6), (2,3), (4,5)
14)(1,6), (2,4), (3,5)
15)(1,6), (2,5), (3,4)
Calling $d(n)$ the number of different draws with $n$ players, I have $d(2)=1$, $d(4)=3$, $d(6)=15$. What is $d(n)$ for $n$ arbitrary even?
