Find a formula for the sum of the squares of the numbers in the $n$th row of Pascal's triangle (i.e. the numbers $\left(\begin{array}{l}n \\ 0\end{array}\right),\left(\begin{array}{l}n \\ 1\end{array}\right), \ldots,\left(\begin{array}{l}n \\ n\end{array}\right)$ ).
We claim that the sum of the squares of the numbers in row $n$ is

$$
\left(\begin{array}{l}
n \\
0
\end{array}\right)^{2}+\left(\begin{array}{c}
n \\
1
\end{array}\right)^{2}+\cdots+\left(\begin{array}{l}
n \\
n
\end{array}\right)^{2}=\left(\begin{array}{c}
2 n \\
n
\end{array}\right)
$$

the middle number in row $2 n$ of Pascal's triangle. To prove this, we can think of the right side as the number of ways to form a committee of $n$ people chosen from some $2 n$ people. For the left side, suppose the $2 n$ people are $n$ men and $n$ women. If there are $k$ women on the committee, then there must be $n-k$ men, so there are $\left(\begin{array}{l}n \\ k\end{array}\right)$ ways to choose the women and $\left(\begin{array}{c}n \\ n-k\end{array}\right)$ ways to choose the men, giving $\left(\begin{array}{c}n \\ k\end{array}\right)\left(\begin{array}{c}n \\ n-k\end{array}\right)=\left(\begin{array}{l}n \\ k\end{array}\right)^{2}$ possible committees with $k$ women. Adding over the possible values of $k$ from 0 to $n$ gives the sum on the left. Since each side counts the same thing in a different way, the two sides must be equal.