Combinatorics proof $\binom{2n}{2}=2\binom{n}{2}+n^2$ The problem is prove that
$$\binom{2n}{2}=2\binom{n}{2}+n^2$$
by showing that each side counts the same collection of subsets.
I am trying to study for a final exam and this is a question from a previous midterm. I literally have no way of going about solving this problem because I've learned absolutely nothing this semester and I only know the formula for n choose k and that's as far as my knowledge goes.
 A: Count the number of ways os choosing a committee of two people from $n$ men and $n$ women. This can be done in two ways.


*

*We need to choose $2$ people from a total of $2n$ people. This can be done in $\dbinom{2n}2$ ways.

*The second way to count is as follows. The committee can be either both men, both women or one man and one woman. Number of ways of forming the committee with both men is $\dbinom{n}2$, number of ways of forming the committee with both women is $\dbinom{n}2$ and the number of ways of forming the committee with one man and one woman is $n^2$.

EDIT In general, if we divide $2n$ into two groups with $i$ and $j$ objects, i.e., $i+j=2n$, by the above argument we have
$$\dbinom{2n}2 = \dbinom{i}2 + \dbinom{j}2 + ij$$
A: Hint
$$(1+x)^{2n}=\big((1+x)^n\big)^2.$$
By using Binomial theorem, you can conclude.
A: In a class, there are $n$ boys and $n$ girls. We want to choose $2$ people from this group. This can be done in $\binom{2n}{2}$ ways. 
Let us count the number of ways to select two people in another way. We could choose $2$ boys, which can be done in $\binom{n}{2}$ ways, or $2$ girls (same), or a boy and a girl. Choosing a boy and a girl can be done in $n^2$ ways.
Remark: Or else you are ordering a pizza. There are $n$ different meat toppings available, and $n$ different vegetarian toppings. In how many ways can you choose a $2$-topping pizza?
