Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$ So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$ 

Here is my solution:

Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from $A$, both from $B$ or one from each. There are ${n \choose 2}$ from $A$ and ${n \choose 2}$ from $B$. For the mixed pair, each choice from $A$ can be coupled with $n$ choices from $B$ so the total is $n^2$. 

Therefore ${2n\choose 2} = 2{n \choose 2} + n^2$ 

Is this correct. Is there any other combinatoric proof to solve this?
Just another way to think about it:

Let's find the coefficient of $x^2$ in the expansion of $(1+x)^{2n}$. On one side, it's equal to ${2n}\choose2$. On the other hand, it's equal to the coefficient of $x^2$ in the expansion of $(1+x)^n(1+x)^n$ which is equal to the sum of products of coefficients of $x^0$ and $x^2$, $x^1$ and $x^1$, and $x^2$ and $x^0$ in the expansion of $(1+x)^n$. That's equal to ${n\choose2}\cdot{ n\choose0}+{n\choose1}\cdot{ n\choose1}+{n\choose0}\cdot{ n\choose2}$ and the result follows.
A more direct (and, in my opinion, simple) approach is to expand and simplify until both sides are equivalent.

$${{2n}\choose{2}}=2{{n}\choose{2}}+n^2$$

First, expand both sides using ${{n}\choose{k}}=\frac{n!}{k!(n-k)!}$:

$$\frac{(2n)!}{2!(2n-2)!}=2\left[{\frac{n!}{2!(n-2)!}}\right]+n^2$$

Expand again, this time focusing on the factorials:

$$\frac{(1)(2)(3)...(2n-2)(2n-1)(2n)}{(1)(2)*(1)(2)(3)...(2n-2)}=2\left[\frac{(1)(2)(3)...(n-2)(n-1)(n)}{(1)(2)*(1)(2)(3)...(n-2)}\right]+n^2$$

In this format, it's easy to see that some variables of the factorials on each side cancel each other out:

$$\frac{(2n-1)(2n)}{2}=(n-1)(n)+n^2$$

Factor in the monomials on each side:

$$\frac{4n^2-2n}{2}=n^2-n+n^2$$

Simplify both sides for the last time:

$$2n^2-n=2n^2-n$$

This method takes a little longer but I feel it's more intuitive and solid.