Combinatorical explanation for $\frac{(2n)!}{2^{n}(n!)}$ being the number of ways to split $2n$ objects into groups of $2$? I am wondering whether there is a simple way to understand why $\frac{(2n)!}{2^{n}n!}$ is equal to the number of ways to split $2n$ objects into groups of $2$? I get exercise questions involving this formula but I want to know why this formula is used. :)
 A: Let's say we already have a grouping of $2n$ objects into pairs. We could write this down as a big list of $2n$ objects, where the first and second items in the list form a pair, the third and fourth are a pair, and so on. However, we have a lot of redundancy in this list, meaning that the same grouping could be written out as many different lists:


*

*For each pair of objects $\{A, B\}$, we could write $A$ onto the list followed by $B$, or $B$ onto the list followed by $A$. So for each pair we have, we get a factor of $2$ to the number of lists we could write.

*The pairs themselves don't come with an ordering, so we could reorder the pairs and write down a different list, which would mean the same grouping of objects into pairs. There are $n!$ ways to rearrange the pairs on the list.


The two facts above mean that a single grouping of objects into pairs could be written in $2^n n!$ different ways as a list of $2n$ objects. There are $(2n)!$ possible lists of $2n$ objects, and so we get the formula.
