<s>[INST] How to prove $\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0$ I would like to prove that: 
\begin{equation*}
\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0;~k\geq0 ; n\geq1. 
\end{equation*}
Can any one help me how to do that? Thanks
 [/INST] Just for fun, here's a generating function answer.  Fix $k \geq 0$ and let $a_n = (-1)^n$ and $b_n = \binom{n}{k}$.  Letting $a$ and $b$ be the exponential generating functions for these sequences, then
\begin{align*}
a(x) = \sum_{n \geq 0} (-1)^n \frac{x^n}{n!} = \sum_{n \geq 0} \frac{(-x)^n}{n!} = e^{-x}
\end{align*}
and
\begin{align*}
b(x) &= \sum_{n \geq 0} \binom{n}{k} \frac{x^n}{n!} = \sum_{n \geq k} \binom{n}{k} \frac{x^n}{n!} = \sum_{n \geq k} \frac{n!}{k!(n-k)!} \frac{x^n}{n!} = \sum_{n \geq k} \frac{1}{k!(n-k)!} x^n\\
&= \frac{x^k}{k!}\sum_{n \geq k} \frac{1}{(n-k)!} x^{n-k} = \frac{x^k}{k!}\sum_{j \geq 0} \frac{1}{j!} x^{j} = \frac{x^k}{k!} e^x
\end{align*}
where we have made the change of index $j = n-k$.  By the convolution formula for exponential generating functions, then
\begin{align*}
\sum_{i=0}^n \binom{n}{i} (-1)^i \binom{n-i}{k} &= \left[\frac{x^n}{n!}\right] a(x) b(x) = \left[\frac{x^n}{n!}\right] e^{-x} \frac{x^k}{k!} e^{x}\\
&= \left[\frac{x^n}{n!}\right] \frac{x^k}{k!} =
\begin{cases}
1 & \text{if } \ n=k\\
0 & \text{otherwise}
\end{cases}
\end{align*}
where $\left[\frac{x^n}{n!}\right]$ is the coefficient of $\frac{x^n}{n!}$ in the formal series.
</s>