Show that $\binom{2n}{n}$ is an even number, for positive integers $n$. I would appreciate if somebody could help me with the following problem

Show by a combinatorial proof that
$$\dbinom{2n}{n}$$
is an even number, where $n$ is a positive integer.

I tried to solve this problem but I can't.

$\binom{2n}{n}=\binom{n}{0}^2+\binom{n}{1}^2+\dots \binom{n}{n}^2\equiv \binom{n}{0}+\binom{n}{1}+\dots \binom{n}{n}=2^n$

$2^n\binom{2n}{n}=\color{red}{\binom{2n}{1}+\binom{2n}{2}+\dots \binom{2n}{n-1}}+\color{blue}{\binom{2n}{n+1}+\binom{2n}{n+2}+\dots\binom{2n}{2n}}=2\binom{2n}{1}+\binom{2n}{2}+\dots \binom{2n}{n-1}$ and hence odd.

The Kneser graph $K(2n,n)$ has $\binom{2n}{n}$ vertices and is $1$-regular. Since the number of odd vertices in any graph is even, $\binom{2n}{n}$ is even.
