Can you elaborate the following combinatorial proof in details?

Question

I know that the number of permutations is always an integer. But I could't understand how the proof is using the fact $n= 2k$ during the proof. This is the proof.

Answer 1

The proof uses $n=2k$ in setting up the $2k$ varibles to permute. Then the argument is by the multinomial theorem.

Answer 2

The proof shows that if $n=2k$ then $n!$ is divisible by $2^k$. It uses a combinatorics argument to establish this result (and clearly the argument uses the fact that $n$ is even, for $n$ is $k$ pairs), but the end result is not about combinations (or permutations) but an elementary result about whole numbers.

Answer 3

You can try another method to prove it. Consider $n$ gifts to be distributed among $k$ people such that each person receives $2$ gifts. (Because we have $n=2k$)

Now for the number of ways. We make groups of 2 gifts each and distribute it among k people. Hence $$(\frac{n!}{2!.2!.2!.....[k times]} . \frac{1}{k!}). k! $$

Which on simplifying gives the number of ways as $$\frac{n!}{2^k}$$ which is always an integer.