Finding a closed formula for $\sum_{r=0}^k {k \choose r}{k \choose r-1}$ I'm looking to find a closed formula for: $\sum_{r=0}^k {k \choose r}{k \choose (r-1)}$. My conjecture is that it is equal to ${2k \choose k-1}$ but I could not prove it. I'm interested in knowing if there's a way to prove this identity (preferably a combinatorial one), but really any would do. my preference is due to the fact that it helps understanding the intuition behind the proof. Thanks a million!
 A: Your conjecture is correct (though the sum probably needs to start with $r=1$).
You can rewrite as
$$\sum_{r=1}^{k} \binom{k}{k-r} \binom{k}{r-1}$$
which has a combinatorial interpretation.
Picking $k-1$ things from $k+k$ things (choose $r-1$ from first $k$, remaining from the second), which is $$\binom{2k}{k-1}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\sum_{r\ =\ 0}^{k}{k \choose r}{k \choose r - 1}}
=\sum_{r\ =\ 0}^{k}{k \choose r}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{k} \over z^{r}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{k}
\sum_{r\ =\ 0}^{k}{k \choose r}\pars{1 \over z}^{r}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{k}\pars{1 + {1 \over z}}^{k}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2k} \over z^{k}}
\,{\dd z \over 2\pi\ic}=\color{#66f}{\large{2k \choose k - 1}}
\end{align}
