Formula for binomial coefficients Does someone know, if the subsequent formula holds for $m \ge n \ge i \ge 1$ and if yes, can give a reference. 
$$\sum_{k=i}^{m-n+i}\binom{k}{i}\binom{m-k}{n-i} = \binom{m+1}{n+1}$$
Thank you very much! 
 A: What you want is equation (5.26) on page 169 of
 Concrete Mathematics (2nd edition) 
 by Ronald Graham, Donald Knuth, and Oren Patashnik.
Corrected: For integers $m,n\geq0$ and integers $\ell,q$ with  $\ell+q\geq 0$
we have
$$\sum_{-q\leq k\leq \ell}{q+k\choose n}{\ell-k\choose m}={\ell+q+1\choose m+n+1}.$$

Let's now substitute your variables $i\geq 0$ and $n-i\geq 0$ in the bottom 
to obtain
$$\sum_{-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$
In fact, since you have assumed $i>0$, we get even more
$$\sum_{i-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$
That's because ${q+k\choose i}=0$ when $q+k<i$.
Redefining the $k$ variable gives
$$\sum_{i\leq k\leq \ell+q}{k\choose i}{\ell+q-k\choose n-i}={\ell+q+1\choose n+1}.$$ 
Letting $m=\ell+q$ gives 
$$\sum_{i\leq k\leq m}{k\choose i}{m-k\choose n-i}={m+1\choose n+1}.$$ 
Is this the same as your sum? Yes! 


*

*If $n=i$, then this is obvious. 

*When $n>i$, then ${m-k\choose n-i}=0$ for $k>m-n+i$ anyway.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = i}^{m - n + i}{k \choose i}{m - k \choose n - i} & =
\sum_{k = 0}^{\infty}{k + i \choose i}{m - k - i \choose n - i} =
\sum_{k = 0}^{\infty}{k + i \choose k}{m - k - i \choose m - k - n}
\\[5mm] & =
\sum_{k = 0}^{\infty}\bracks{{-i - 1 \choose k}\pars{-1}^{k}}
\bracks{{i - n - 1\choose m - k - n}\pars{-1}^{m - k - n}}
\\[5mm] & =
\pars{-1}^{m - n}\sum_{k = 0}^{\infty}{-i - 1 \choose k}
\bracks{z^{m - k - n}}\pars{1 + z}^{i - n - 1}
\\[5mm] & =
\pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{i - n - 1}\sum_{k = 0}^{\infty}{-i - 1 \choose k}z^{k}
\\[5mm] & =
\pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{i - n - 1}\,
\pars{1 + z}^{-i - 1} =
\pars{-1}^{m - n}\bracks{z^{m - n}}\pars{1 + z}^{-n - 2}
\\[5mm] & =
\pars{-1}^{m - n}{-n - 2 \choose m - n} =
\pars{-1}^{m - n}\bracks{{m + 1 \choose m - n}\pars{-1}^{m - n}} =
\bbx{m + 1 \choose n + 1}
\end{align}
