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!
On
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.
On
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.
On
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
On
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.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}