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.

You should be familiar with the recursive definition of the binomial coefficients, that $\binom{n}{r} = \binom{n-1}{r-1}+\binom{n-1}{r}$
\begin{matrix}1\\
1 & 1\\
1 & 2 & 1\\
1 & 3 & 3 & 1\\
1 & 4 & 6 & 4 & 1\\
\vdots & & \vdots & & \vdots&\ddots\\
 & \cdots& \binom{n-1}{r-1} & \binom{n-1}{r} & \cdots\\
 & \cdots& & \binom{n}{r} & \cdots\end{matrix}
Now, notice that $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$
Notice further that $\binom{2n-1}{n-1} = \binom{2n-1}{(2n-1)-(n-1)} = \binom{2n-1}{n}$
So, $\binom{2n}{n} = 2\cdot \binom{2n-1}{n}$ and is therefore even.

The combinatorial proof of the recursive definition is that to choose $r$ objects out of $n$ with one special object, either the special object is in the choosing or it is not.  Breaking it into cases, if it is, you still need to choose $r-1$ objects out of the $n-1$ remaining.  If it is not, then you still need to choose $r$ objects out of the $n-1$ remaining.
Hence $\binom{n}{r} = \binom{n-1}{r-1}+\binom{n-1}{r}$
