Compute $$\sum_{n=0}^{10}(-1)^n\binom{10}{n}\binom{12+n}{n}.$$
How do I calculate it using well known identities?I've tried more or less every identity I know with alternating signs but I end up with a mess.
(The answer is $66$)
I saw this identity $$\sum_{n=0}^k(-1)^n\binom{k}{n}\binom{m+n}{n}=\binom{m}{k}.$$
But I don't know how to prove it in an easy way.
Actually, the identity should be $$\sum_{n=0}^k(-1)^n\binom kn\binom{m+n}n=(-1)^k\binom mk.$$
First rewrite $\binom{m+n}n=\binom{m+n}m$. Perhaps this is not the pure manipulation of previous identities you may have had in mind, but one way to compute the sum is to use generating functions. Introduce the function $$f(z)=\sum_k\sum_n(-1)^n\binom kn\binom{m+n}mz^k.$$ If we can prove that $f(z)=(1-z)^m$, then we are done because the coefficient of $z^k$ in $(1-z)^m$ is $(-1)^k\binom mk$, which must be the same as the sum we are trying to compute.
To simplify the given definition of $f$, we use the identity $$\frac1{(1-z)^{m+1}}=\sum_i\binom{m+i}mz^i,$$ as well as the equivalent form $$\frac{z^m}{(1-z)^{m+1}}=\sum_i\binom imz^i.$$ So \begin{align*} f(z)&=\sum_k\sum_n(-1)^n\binom kn\binom{m+n}mz^k\\ &=\sum_n(-1)^n\binom{m+n}m\sum_k\binom knz^k\\ &=\sum_n(-1)^n\binom{m+n}m\frac{z^n}{(1-z)^{n+1}}\\ &=\frac1{1-z}\sum_n\binom{m+n}m\left(-\frac z{1-z}\right)^n\\ &=\frac1{1-z}\frac1{\left(1-\Bigl(-\frac z{1-z}\Bigr)\right)^{m+1}}\\ &=(1-z)^m \end{align*} which is what we wanted to prove.
Another easy way to prove the identity is \begin{align*} {\color{blue}{\sum_{n=0}^k(-1)^n\binom{k}{n}\binom{m+n}{n}}}&=\sum_{n=0}^k\binom{k}{k-n}\binom{-m-1}{n}\tag{1}\\ &=\binom{k-m-1}{k}\tag{2}\\ &\,\,\color{blue}{=\binom{m}{k}(-1)^k}\tag{3} \end{align*}
Comment:
In (1) we use the binomial identities $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (2) we use the Chu-Vandermonde identity.
In (3) we use the right-hand identity from (1) again.
Snake oil: Let $[x^k] f(x)$ be the coefficient near $x^k$ in the expansion of $f(x)$.
Use the identities $\binom{n}{k}=[x^k](1+x)^n$ and $[x^a] x^{a-b} f(x) = [x^b] f(x)$:
$$\begin{align} \sum\limits_{n=0}^{k} (-1)^n \binom{k}{n} \binom{m+n}{n} &= \sum\limits_{n=0}^k (-1)^n \binom{k}{n} [x^k] x^{k-n}(1+x)^{m+n} \\&= [x^k] (1+x)^m \sum\limits_{n=0}^k (-1)^n \binom{k}{n} x^{k-n} (1+x)^n \\&= [x^k](1+x)^m(x-(1+x))^k = (-1)^k[x^k](1+x)^m = (-1)^k\binom{m}{k}. \end{align}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{n = 0}^{10}\pars{-1}^{n}\, {10 \choose n}{12 + n \choose n}} \\[5mm] = & \ \sum_{n = 0}^{10}\pars{-1}^{n}\, {10 \choose n}\bracks{{-12 - n + n - 1 \choose n}\pars{-1}^{n}} \\[5mm] = & \ \sum_{n = 0}^{10}{10 \choose n}{-13 \choose -13 - n} \\[5mm] = & \ \sum_{n = 0}^{10}{10 \choose n}\bracks{z^{-13 - n}\,} \pars{1 + z}^{-13} \\[5mm] = & \ \bracks{z^{-13}}\pars{1 + z}^{-13}\, \sum_{n = 0}^{10}{10 \choose n}z^{n} \\[5mm] = & \ \bracks{z^{-13}}\pars{1 + z}^{-13}\, \pars{1 + z}^{10} \\[5mm] = & \ \bracks{z^{-13}}\pars{1 + z}^{-3} = {-3 \choose -13} \\[5mm] = & \ \pars{-1}^{\pars{-3}\ -\ \pars{-13}}\,\,\,\, {-\bracks{-13} - 1 \choose -3 - \bracks{-13}} \\[5mm] = & \ {12 \choose 10} = {12 \times 11 \over 2} = \bbx{\color{#44f}{66}} \end{align}
