Interpretation of a combinatorial expression

Question

I came across this expression :

$ \frac {(2n)!} {2^n n!} $

I know that this evaluates to an odd integer. More specifically :

$ \frac {(2n)!} {2^n n!} = (2n -1) \cdot (2n - 3)\cdots 5\cdot3\cdot1$

I want to know where this expression comes in a real-life scenario. For example $\sum_{k=0}^{n}{n\choose k}$ represents the number of ways to choose a subset from a set of cardinality $n$.

Answer 1

$(2n-1)!!=(2n-1)\cdot(2n-3)\cdot(2n-5)\cdots 5\cdot 3\cdot 1$ is an answer to various combinatorial questions such as the following:

In how many ways can $2n$ distinct people split up into $n$ unlabelled teams each of size two such that everyone is on exactly one team each.

To see the answer, place an arbitrary order on the $2n$ people. Without loss of generality, let the order we use be age.

Take whoever is youngest in the group. Pick who will be partnered with the youngest person. Remove both people from the pool of available people.

Take whoever is youngest from those remaining. Pick who will be partnered with that person. Remove both people from the pool of available people.

Repeat the previous step until everyone is paired off.