We are organizing a tennis tournament with 16 players: each player faces exactly one other player. In how many ways can the $2n$ players be paired (unordered pairs)?
We put the players online, then we choose to put players $2 k+1$ and $2 k+2$ together for $0 \leq k \leq n$. However, we have counted each configuration $2^{n} n!$ times: indeed, for each configuration, we can permute the $n$ pairs, and swap the 2 elements of each pair. Thus, the result is $\frac{(2 n)!}{n!2^{n}}=1 \times 3 \times \cdots \times(2 n-3) \times(2 n-1)$