TITLE: Sum of binomial coefficients $\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0$
QUESTION [18 upvotes]: How do I prove the following identity:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0$$
I am trying to use inclusion-exclusion, and this will boil down to a sum like inclusion-exclusion, and the $\binom{2n-2k}{n-1}$ term wouldn't matter (it will be equivalent to set sizes). Is this a correct way to go?

REPLY [4 votes]: Another way to do this is to use snake oil. Use $m$ as a new free variable (for $n - 1$), notice that outside the given range the first binomial coefficient is zero, and write the generating function for the sum we want:
$\begin{align}
\sum_{m \ge 0} z^m \sum_k (-1)^k \binom{n}{k} \binom{2 n - 2 k}{m}
  &= \sum_k (-1)^k \binom{n}{k} \sum_{m \ge 0} \binom{2 n - 2 k}{m} z^m \\
  &= \sum_k (-1)^k \binom{n}{k} (1 + z)^{2 n - 2 k} \\
  &= (1 + z)^{2 n} \sum_k (-1)^k \binom{n}{k} (1 + z)^{- 2 k} \\
  &= (1 + z)^{2 n} \left(1 - (1 + z)^{-2}\right)^n \\
  &= \left((1 + z)^2 - 1\right)^n \\
  &= z^n (2 + z)^n
\end{align}$
Now we want the coefficient of $z^m$ of this:
$\begin{align}
[z^m] z^n (2 + z)^n
  &= [z^{m - n}] (2 + z)^n \\
  &= \binom{n}{m - n} 2^{n - (m - n)} \\
  &= \binom{n}{m - n} 2^{2 n - m}
\end{align}$
The original problem asks for $m = n - 1$, sure enough, $\binom{n}{-1} = 0$.
Note: The $[z^n]$ notation is described by Knuth as a simpler way to handle Egorychev's "method of coefficients", which I believe is what Marko Riedel is doing.