242. Sum of squares of binomial coefficients. Find the sum of the squares of the coefficients in the expansion of $(a+b)^{n}$.
242. If we select $n$ balls at random from $n$ red and $n$ black balls, the number of possible combinations is

$$
\begin{gathered}
C_{2 n}^{n}=C_{n}^{0} \cdot C_{n}^{n}+C_{n}^{1} \cdot C_{n}^{n-1}+C_{n}^{2} \cdot C_{n}^{n-2}+\ldots \\
\ldots+C_{n}^{n-1} \cdot C_{n}^{1}+C_{n}^{n} \cdot C_{n}^{0}
\end{gathered}
$$

But $C_{n}^{k}=C_{n}^{n-k}$, so

$$
C_{2 n}^{n}=\sum_{i=0}^{n}\left(C_{n}^{i}\right)^{2}
$$

[M. M., 24, 54 (September 1950).]

This is equivalent to the statement that the sum of the squares of the numbers on the $k$-th diagonal of Pascal's triangle (see problem 179) is equal to the $k$-th (middle) number from the $(2 k-1)$-th diagonal. Another short proof of this relation was given by G. Sziller.

$[A . M . M ., 42,46$ (January 1935).]