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
 A: 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.
A: HINT:
$$\binom{n}{i}\binom{n-i}{k}=\dfrac{n!}{i!\,(n-i)!}\cdot\dfrac{(n-i)!}{k!\,(n-i-k)!}=\dfrac{n!}{i!\,k!\,(n-i-k)!}=\binom{n}{k}\binom{n-k}{i}$$
A: Here is a  solution using complex variables that  involves only finite
sums.

Suppose we seek to evaluate
$$S_k(n) = \sum_{q=0}^n (-1)^q {n\choose q} {n-q\choose k}.$$
Introduce
$${n-q\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{k+1}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\sum_{q=0}^n {n\choose q} (-1)^q 
\frac{(1+z)^{n-q}}{z^{k+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{k+1}}
\sum_{q=0}^n {n\choose q} (-1)^q 
\frac{1}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{k+1}}
\left(1-\frac{1}{1+z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{k+1}}
\frac{z^n}{(1+z)^n} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1-n}} \; dz.$$
This is zero by inspection except when $k+1-n=1$ or $k=n$
when it evaluates to one.
A: Corrected HINT: For $k=0$ you can verify it by direct calculation; note that here you really do need $n>0$. As noted in the comments, the result is false for $k=n$. For $k>0$ show that 
$$\sum_{i=0}^n(-1)^i\binom{n}i\binom{n-i}k$$
is an inclusion-exclusion calculation of the number of subsets of $[n]=\{1,2,\ldots,n\}$ of size $k$ that contain every element of $[n]$; since there are clearly no such sets when $k<n$, this will prove that in those cases the sum is $0$.
A: Note that only when $n \neq k$, the equation holds.
When $n = k$, $\sum_{i=0}^{n}(-1)^i{n \choose i}{n - i \choose k} = {n \choose 0}{n \choose n} = 1$.
When $n > k$, besides Demosthene's hint, I give another hint below. 
$$
(1 - x)^{n-k} = \sum_{i=0}^{n-k}{n -k \choose i}(-1)^i x^i
$$
The case for $n < k$ is trivial.
