How can you prove $\binom {2n}{2} = 2\binom{n}{2} + n^2$? Is there a way one can prove that this is true: ${2n \choose 2} = 2{n\choose 2} + n^2$ ?
I am thinking it may involve binomial theorem. 
 A: We have 
$$\begin{align}
\binom{2n}{2}&=\frac{(2n)!}{2!(2n-2)!}\\\\
&=n(2n-1) \tag 1
\end{align}$$
and
$$\begin{align}
2\binom{n}{2}+n^2&=2\frac{n!}{2!(n-2)!}\\\\
&=n(n-1)+n^2\\\\
&=n(2n-1) \tag 2
\end{align}$$
Upon comparing the right-hand sides of $(1)$ and $(2)$ we obtain the result!
A: Combinatorial method: consider the following diagram.

The number of lattice points in the big square, such that the first coordinate is strictly less than the second, is equal to those in the top-left little square, plus the two little triangles.
A: Suppose we wish to select two people from a group of $n$ men and $n$ women.  Then $\binom{2n}{2}$ is the number of ways we can select two people from the group.  Alternatively, we can select two men in $\binom{n}{2}$ ways, two women in $\binom{n}{2}$ ways, and one man and one woman in $n \cdot n = n^2$ ways.  Hence, $$\binom{2n}{2} = \binom{n}{2} + \binom{n}{2} + n \cdot n = 2\binom{n}{2} + n^2$$
A: Induction is fine, too. We have: $$\binom{a}{2}=\binom{a-1}{2}+\binom{a-1}{1},$$ hence we just need to check that the given identity holds for $n=1$ and that $n^2-(n-1)^2=2n-1$.
A: You have collection of $n$ red and $n$ black balls. In how many ways we can choose 2 balls form this set?
First solution: ${2n}\choose{2}$.
Second solution: we can choose two red or two black balls in $2 {{n}\choose{2}}$ ways and a pair of one red and one black ball in $n \cdot n = n^2$ ways.
Hence ${{2n}\choose{2}} = 2 {{n}\choose {2}} + n^2$.
