Proof of a Binomial Identity using a combinatorial argument: $\binom{2k}{2} =\binom{k−l}{2}+ \binom{k+l}{2}+ k^2 − l^2$ Question 
Prove that if $k$ and $l$ are two positive integer with $k ≥ l$, then $\binom{2k}{2} =\binom{k−l}{2}+ \binom{k+l}{2}+ k^2 − l^2$
using a combinatorial argument.
I tried using Vandermonde's Identity and Pascal's Identity but that's leading me nowhere. 
Any lead is very welcome. 
Thanks
 A: Hint: The left-hand side is the number of ways to choose $2$ items out of $2k$. The right-hand side can be interpreted as the number of ways to choose $2$ items out of one subset, two items out of the remaining subset, and one item from each (using $(a+b)(a-b)=a^2-b^2$).
A: Suppose you want to pick $2$ elements from a set of $2k$ elements. Color $k-l$ of the elements blue, and the remaining $k+l$ elements red. Then you can either pick $2$ blue elements (in ${k-l} \choose 2$ ways), or $2$ red elements (in ${k+l} \choose 2$ ways), or you can pick one blue element and one red element (in $(k-l)(k+l)=k^2-l^2$ ways).
A: You want a combinatorial argument...
Say you have $2k$ items and you want to select $2$ of them. Divide the $2k$ items into two groups, one with $k+\ell$ items and one with $k-\ell$ items.
To select two items out of the overall $2k$, you could select two from the group of $k+\ell$ items; or two from the group of $k-\ell$ items; or you could select one from the group of $k+\ell$ and one from the group of $k-\ell$ items. If you add all three possibilities, you should get $\binom{2k}{2}$. 
How many ways to pick two from $k+\ell$? How many ways to pick two from $k-\ell$? How many ways to pick one from $k+\ell$ and one from $k-\ell$?
