# Solving Binomial Coefficients with Double Counting

I have a problem that I am trying to solve two different ways.

The problem is:

The following equality holds, for a positive integer $n$: $$\dbinom{2n}{2} = 2\dbinom{n}{2} + n^2$$

Show that it does by double counting and by algebraic manipulation.

I am assuming that the problem is asking me to show that both sides of the equality are equal to each other (this area of math is new to me). If that is the case, I did the algebraic portion using the absorption property . What I cannot figure out is how to prove the equality using double counting. Can someone point me in the right direction and help me figure this out?

EDIT: Thanks for formatting!

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We have $n$ boys and $n$ girls, and want to choose two people. There are $\binom{2n}{2}$ ways to do this.
Let us count another way. We can choose $2$ boys, or $2$ girls, or one of each. There are $\binom{n}{2}$ ways to choose $2$ boys, and $\binom{n}{2}$ ways to choose $2$ girls, and $\binom{n}{1}\binom{n}{1}=n^2$ ways to choose one of each.
Remark: Note that the same argument gives $\binom{m+n}{2}=\binom{m}{2}+\binom{n}{2}+mn$. One can generalize further.