Is this a proper combinatorial proof? Give a combinatorial proof of the following identity for every positive integer $n\ge1$:
$$\binom{2n}2=2 \cdot \binom{n}2 + n^2$$
The first part can be seen as picking $2$ elements out of $2n$ elements.
The second part can then be seen as picking $2$ elements out of $n$ + picking $2$ elements out of the remaining $2n-n=n$ + picking $1$ elements from the first $n$ elements and $1$ from the second $n$ elements ($n$ choices for the first and $n$ choices for the second).
Which comes down to $\binom{n}2 + \binom{n}2 + n \cdot n=2 \cdot \binom{n}2 + n^2$
So i'm not interested in the algebraic way, I can do that.
What I would like to know is, does this proof my statement and is it a proper combinatorial proof.
Many thanks!
 A: The basic idea is there, but it lacks the crucial final step of explaining clearly how your interpretation of the righthand side amounts to the same thing as picking a pair of elements from a set of $2n$ elements. You could do that as follows.

Let $S$ be a set of $n$ couples, for a total of $2n$ people. There are $\binom{2n}2$ ways to choose two of these people. Alternatively, we can count the possible pairs according to sex: there are $\binom{n}2$ ways to choose a pair of men, another $\binom{n}2$ ways to choose a pair of women, and $n^2$ ways to choose a mixed pair. Thus, $\binom{2n}2=2\binom{n}2+n^2$.

A: I feel there's some potential for stating it more clearly, but yes, that's a proper combinatorial proof.
A: Just a slight modification of your own answer: RHS can be represented as 
$$
\binom{2}{1} \cdot \binom{n}{2} + \binom{n}{1}\cdot\binom{n}{1}
$$
which can be interpreted as either two items out of $n$ AND one item out of these 2 OR divide the original set into two equal subsets AND select one item out of each subset. 
A: $\binom{2n}2$ is equal all of subsets that have $2$ element from a set with $2n$ elements. then you can first divide this $2n$ element as $n$ group and other $n$ group and you can form one special group choose two element!(it has $2*\binom{n}2$ way).or you can choose from each group only one element (it has $n*n$ way). so you now have all of subset (with $2$ element) from a set with $2n$ element.and so your equal proved!
