Prove that $\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m+k}{r}=0$ It is well-known that $\sum_{k=0}^n(-1)^k\binom{n}{k}=(1-1)^n=0$.
It is seems like that
$$\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m+k}{r}=0$$ 
for any $m,r\in\mathbb{N}$, $r\leq m$.
How to prove or disprove this?
Thank you.
 A: $$
\begin{align}
\sum_k(-1)^k\binom{n}{k}\binom{m+k}{r}
&=\sum_k(-1)^k\binom{n}{k}\sum_j\binom{k}{j}\binom{m}{r-j}\tag{1}\\
&=\sum_j\binom{m}{r-j}\sum_k(-1)^k\binom{n}{k}\binom{k}{j}\tag{2}\\
&=\sum_j\binom{m}{r-j}(-1)^n[n=j]\tag{3}\\
&=(-1)^n\binom{m}{r-n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: Vandermonde's Identity
$(2)$: change order of summation
$(3)$: identity $(1)$ shown in this answer (Inverse Binomial Transform)
$(4)$: apply Iverson Brackets
This is $0$ if $n\gt r$.
A: I think you want $r<n$. Then $\binom{m+k}{r}$ is a polynomial in $k$ of degree $r$. (Easy to see by using the definition.) Now, the general result is
$$ \sum_{k=0}^n (-1)^k \binom{n}{k} P(k) = 0, $$
if $P$ is any polynomial of degree less than $n$. (The simplest way to prove this is to apply the differential operator $P(x \frac{d}{dx})$ to $(1-x)^n$.) Therefore if $r<n$, we get zero for your sum.
A: Hints:


*

*$p(m)=\binom{m}{r}$ is a polynomial with degree $r$;

*if $\delta$ is the difference operator that maps a polynomial $q(x)$ into $q(x)-q(x+1)$ and the degree of $p(x)$ is $d\geq 1$, the degree of $(\delta p)(x)$ is $d-1$;

*$(\delta^2 p)(x)= p(x)-2p(x+1)+p(x+2)$ and $(\delta^3 p)(x)=p(x)-3p(x+1)+3p(x+2)-p(x+3)$, so:
$$ \sum_{k=0}^{n}(-1)^k \binom{n}{k}\binom{m+k}{r} = (\delta^n p)(m). $$
So we have that $\color{red}{n>r}$ ensures that our sum is zero. If $n=r$ our sum equals $(-1)^n$.
A: This exercise is straightforward.
Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k {n\choose k} {m+k\choose r}.$$
Introduce
$${m+k\choose r} = 
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m+k} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m} 
\sum_{k=0}^n (-1)^k {n\choose k} (1+z)^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m} 
(1-(1+z))^n
\; dz
\\ = \frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r-n+1}} (1+z)^{m} 
\; dz.$$
This evaluates by inspection to
$$(-1)^n {m\choose r-n}.$$
A: A little preliminary algebra allows an easy combinatorial argument. Let
$$f(m,n,r)=\sum_{k=0}^n(-1)^k\binom{n}k\binom{m+k}r\;;$$
then
$$\begin{align*}
(-1)^nf(m,n,r)&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{n-k}\binom{m+k}r\\
&=\sum_{k=0}^n(-1)^k\binom{n}k\binom{m+n-k}r\;.
\end{align*}$$
Now suppose that you have $m$ men and $n$ women, and you want to know in how many ways you can choose a committee of $r$ from these $m+n$ people in such a way that it includes every woman. On the one hand this is simply the number of ways to choose $r-n$ men for the committee, which is clearly $\binom{m}{r-n}$. 
On the other hand, for any subset $S$ of the set $W$ of women let $\mathscr{C}_W$ be the set of $r$-person committees that do not include any of the women in $S$; clearly
$$\left|\mathscr{C}_S\right|=\binom{m+n-|S|}r\;.$$
For $0\le k\le n$ there are $\binom{n}k$ sets of $k$ women, so
$$\left|\bigcup_{w\in W}\mathscr{C}_{\{w\}}\right|=\sum_{k=1}^n(-1)^{k-1}\binom{n}k\binom{m+n-k}r$$
by the inclusion-exclusion principle, and the number of $r$-person committees containing every woman is
$$\begin{align*}
\binom{m+n}r-\sum_{k=1}^n(-1)^{k-1}\binom{n}k\binom{m+n-k}r&=\binom{n}0\binom{m+n}r+\sum_{k=1}^n(-1)^k\binom{n}k\binom{m+n-k}r\\
&=\sum_{k=0}^n(-1)^k\binom{n}k\binom{m+n-k}r\\
&=(-1)^nf(m,n,r)\;.
\end{align*}$$
Thus
$$(-1)^nf(m,n,r)=\binom{m}{r-n}\;,$$
and hence
$$f(m,n,r)=(-1)^n\binom{m}{r-n}\;,$$
which is $0$ ifr $r<n$.
