I came across this problem and this related problem while browsing through the combinatorics section. Although both of them have been closed, I find the discussion inside the solutions inside the posts quite interesting and wish to reinitiate the discussion and seek for alternative solutions if any.
Problem Statement
Both authors wished to prove the following:
$$ \binom{2n}{2}=2\cdot\binom{n}{2}+n^{2} $$
Sample Solution from The Posts
User N.F. Taussig argued that if one wishes to select $2$ people out of $n$ men and $n$ women, one can:
- Directly select two out of everyone in $\binom{2n}{2}$ ways, or
- Select two out of all men, two out of all women, or one of each in $\binom{n}{2}$, $\binom{n}{2}$, $n^{2}$ ways respectively and $2\cdot\binom{n}{2}+n^{2}$ in total
In conclusion, both LHS and RHS count the same object and must be equal.
In a separate answer, Graham Kemp suggest to find a counting argument for $\binom{2n}{2}=\binom{2}{1}\binom{n}{2}+n^{2}$
My Solution
Through simple modification to the RHS, the problem seems to be much easier:
$$ \binom{2n}{2}=\binom{n}{2}\binom{n}{0}+\binom{n}{1}\binom{n}{1}+\binom{n}{0}\binom{n}{2} $$
One can immediately notice that this is nothing but the Vandermonde's identity. In this form, Taussig's combinatorial argument also become really obvious.
Remark
I find it interesting because two of the most basic expressions in combinatorics i.e. $\binom{n}{0}=1$ and $\binom{n}{1}=n$ can make such combinatoric problem trivial. I also with to know if there are other alternative solutions.
You can define a function $$f(n) = {2n\choose 2} -2{n\choose 2}-n^2$$
Clearly it is a polynomial of a degree 2, so in other to prove $f(n)=0$ for all $n$ it is enough to prove for $3$ natural numbers, say $n=1,2,3$ and you are done.