Combinatorical proof of the identity $\binom{2n}{2} = 2\binom{n}{2}+n^2$

So, the given expression is $$\binom{2n}{2} = 2\binom{n}{2}+n^2$$

The task is to give a combinatorical proof for it.

Left side of the identity is obviously equal to the number of options for choosing 2 elements out of the set with cardinality $2n$.

What issues me is that I can't think of any way to separate that into two disjoint cases which would have $2\binom{n}{2}$ and $n^2$ different options (what is, I believe, meant to happen).

Hint: split the set into two halves and consider whether the selected elements are in the same half.

Consider 2 sets:

$A=${$a_1,a_2,....a_n$}, and

$B=${$b_1,b_2,..., b_n$}, all distinct elements.

LHS:

The number of ways to choose $2$ elements from $A\cup B$ :

$\binom{2n}{2}.$

RHS:

Choose $2$ elements from $A$, or from $B$ :

In $2\binom{n}{2}$ ways

The mix:

$S_{i,k} =${$a_i,b_k$} , $1\le i,k \le n$.

How many different sets $S_{ik}$?

$n$ ways to chose from $A$, and $n$ ways to choose from $B$:

Altogether: $n^2$ ways.