Find the sum of series: $\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n}$ To find the sum:
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n}$$
Try:
I do not have any clue about the question. I was thinking of finding coefficient of some required power in a binomial expansion, but wasn't able to proceed as the power of $x$ seems to be non-constant in each term ($x^{n+k}$).
Please give a small hint!
 A: Note that
\begin{align*}
\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n}&=\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n-k}\\
&=\sum_{k=0}^{n} (-1)^k \binom{n}{k} (-1)^{n-k}\binom{-(n+1)}{n-k}\\
&=(-1)^{n}\sum_{k=0}^{n} \binom{n}{k} \binom{-(n+1)}{n-k}\\
&=(-1)^{n}\binom{n-(n+1)}{n}=1.
\end{align*}
where we used 
$$\binom{2n-k}{n-k}=\frac{(2n-k)\cdots(n+1)}{(n-k)!}=
(-1)^{n-k}\frac{(-n-1)\cdots(-2n+k)}{(n-k)!}=
(-1)^{n-k}\binom{-n-1}{n-k}$$
and the Vandermonde's identity.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{2n - k \choose n} & =
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{z^{n}}\pars{1 + z}^{2n - k}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{2n}
\sum_{k = 0}^{n}{n \choose k}\pars{-\,{1 \over 1 + z}}^{k}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{2n}\pars{1 - {1 \over 1 + z}}^{n}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{n}\,z^{n} = \bbx{\large 1} \\ &
\end{align}
A: Another way is to exploit the Melzak's identity: $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k $$ where $f $ is an algebraic polynomial up to degree $n $. So taking $f\left(z\right)=\dbinom{z+n}{n}z$ we get $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\dbinom{y+n-k}{n}\frac{y-k}{-x-k}=-\frac{\dbinom{n+y+x}{n}}{x\dbinom{x+n}{n}}\left(x+y\right)$$ so taking $y=n$ and the limit $x\rightarrow-n$ we get $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\dbinom{2n-k}{n}=-\lim_{x\rightarrow-n}\frac{\dbinom{2n+x}{n}}{x\dbinom{x+n}{n}}\left(n+x\right)=\color{red}{1}$$ as wanted.
A: I think that the simpler  and shorter way would be:
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,k} \left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  2n - k \cr 
  n \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,k\, \le \,n} {\left( \matrix{
  k - n - 1 \cr 
  k \cr}  \right)\left( \matrix{
  2n - k \cr 
  n \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,k\, \le \,n} {\left( \matrix{
  k - n - 1 \cr 
  k \cr}  \right)\left( \matrix{
  2n - k \cr 
  n - k \cr}  \right)}  =   \cr 
  &  = \left( \matrix{
  n \cr 
  n \cr}  \right) = 1 \cr} 
$$
 where


*

*1st step : Upper Negation $  \left({ - 1} \right)^{\,k} \left( \matrix{
  n \cr 
  k \cr}\right)=\left( \matrix{
  {k-n-1} \cr 
  k \cr}\right)$

*2nd step: Symmetry $\left( \matrix{
  n \cr 
  k \cr}\right)=\left( \matrix{
  n \cr 
  {n-k} \cr}\right)\quad |0 \le \text{integer} \,n$

*3rd step: Double Convolution 
$$
\sum\limits_{a\, \le \,k\, \le \,b} {\left( \matrix{
  k - c \cr 
  k - a \cr}  \right)\left( \matrix{
  d - k \cr 
  b - k \cr}  \right)}  = \left( \matrix{
  d - c + 1 \cr 
  b - a \cr}  \right)
$$

A: There is also a combinatorial solution. Consider the following problem:

How many subsets of $\{1,2,\dots,2n\}$ of size $n$ contain none of the numbers in $\{1,2,\dots,n\}$?

On the one hand, the answer is obviously $1$; the only such subset is $\{n+1,n+2,\dots, 2n\}$. 
On the other hand, we can solve this with the principle of inclusion exclusion. Letting $E_i$ be the number of subsets of size $n$ which do contain $i$, we need to count $E_1^\complement\cap E_2^\complement\cap \dots \cap E_n^\complement$. For any $k$ indices $i_1,\dots,i_k$, the size of the intersection $E_{i_1}\cap E_{i_2}\cap \dots \cap E_{i_k}$ is $\binom{2n-k}n$. Taking the alternating sum over all such subsets of indices, we get your sum.
A: This is inspired by the answer of @Marco Cantarini that noticed that the expression
$$k \mapsto \binom{2n-k}{n}= \colon P(k)$$
is a polynomial $P$ in $k$ of degree $n$ with leading coefficient $\frac{(-1)^n}{n!}$. The sum therefore equals $(-1)^n (\Delta^n P )(0) = (-1)^n \cdot n! \cdot \frac{(-1)^n}{n!}= 1$.
Details: we know that if $f(x)$ is a polynomial in $x$ of degree $n$ with leading term $a$ then $\Delta f(x) \colon =f(x+1) - f(x)$ is a polynomial of degree $n-1$ and leading term $n \cdot a$. We conclude that $\Delta^n f$ is a constant polynomial  $n! a$. Note that we have
$$\Delta^n f(x) = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} f(x+k)$$
Now take $x=0$ and the initial polynomial.
