This is rather well known identity: If each of m, n, & r is a positive integer and r exceeds neither m nor n then \(\displaystyle \L
\left( \begin{array}{c}
m + n \\
r \\
\end{array} \right) = \sum\limits_{k = 0}^r {\left( \begin{array}{l}
m \\
k \\
\end{array} \right)\left( \begin{array}{c}
n \\
r - k \\
\end{array} \right)}\) .
To see why this is true, suppose that each of A & B is a set with no elements in common and A has m elements and B has n elements. The union of A with B has m+n elements and we can choose r of those in \(\displaystyle \L
\left( \begin{array}{c}
m + n \\
r \\
\end{array} \right)\) ways.
On the other hand, for each \(\displaystyle \L
0 \le k \le r\) that is the same as if we choose k from A and r-k from B.
If we allow m=n=r the we get \(\displaystyle \L
\left( \begin{array}{c}
2n \\
n \\
\end{array} \right) = \sum\limits_{k = 0}^r {\left( \begin{array}{l}
n \\
k \\
\end{array} \right)^2 }\) noting that \(\displaystyle \L
\left( \begin{array}{l}
n \\
k \\
\end{array} \right) = \left( \begin{array}{c}
n \\
n - k \\
\end{array} \right)\).
I hope this helps and is not too late.
