How to give a combinatorial proof for: If $n$ and $k$ are positive integers with $n=2k$ then $\frac{n!}{2^k}$ is an integer

Question

How can i give a combinatorial proof for if $n$ and $k$ are positive integers with $n=2k$ then $\dfrac{n!}{2^k}$ is an integer?

Answer 1

Suppose you have $k$ pairs where the objects in a given pair are identical but the objects in any two pairs are distinct. That is to say, you have two of object $a$, two of object $b$, and so on down to two of object $k$.

Thus you have $n=2k$ objects all told and every object has a unique duplicate.

Now the number of ways to arrange those $n$ objects in a line is precisely your quotient, which is therefore an integer.

Answer 2

The permutations of the socks from $k$ pairs can be obtained as all permutations with the socks of every pairs considered identical, and all these permutations can be duplicated by swapping the socks of the pairs.

E.g.

$ABBA$ yields the four $ABba,aBbA,ABba,abBA$.

Answer 3

If $n=2k$, $\dfrac{n!}{2^k}$ can be written as the multinomial $\dbinom{n}{2,\dots,2}$ (with $k$ $2$'s), and is therefore an integer.