TITLE: How many integer solutions are there of the equation $|x_{1}|+|x_{2}|+\cdots +|x_{k}|=n$?
QUESTION [7 upvotes]: How many solutions are there to the equation
$$|x_{1}|+|x_{2}|+\cdots +|x_{k}|=n$$
for $n,k\in \mathbb N$ and $\forall\ 1\leq i\leq k,\ x_{i}\in \mathbb Z$?

Any ideas? I don't know how to approach this question. (all I know that if my  $ x_{i}$'s were in $\mathbb N$ the solution would be $\binom{n+k-1}{k-1}$  )

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\delta_{\ell\ell'}}$ is the Kronecker Delta which is equal to $\ds{1}$ if $\ds{\ell = \ell'}$ and $\ds{0}$ otherwise. The following multiple sum 'counts' $\ds{1}$ each time $\ds{\verts{x_{1}} + \cdots + \verts{x_{k}} = n}$.
  So,

\begin{align}
&\color{#f00}{\sum_{x_{1} = -\infty}^{\infty}\cdots
\sum_{x_{k} = -\infty}^{\infty}
\delta_{\verts{x_{1}} + \cdots + \verts{x_{k}}\,,\,n}} =
\sum_{x_{i} = -\infty}^{\infty}\cdots\sum_{x_{k} = -\infty}^{\infty}\
\overbrace{%
\oint_{\verts{z} = 1^{-}}{1 \over z^{n + 1 - \verts{x_{1}} - \cdots - \verts{x_{k}}}}\,{\dd z \over 2\pi\ic}}^{\ds{\delta_{\verts{x_{1}} + \cdots + \verts{x_{k}}\,,\,n}}}
\\[3mm] = &\
\oint_{\verts{z} = 1^{-}}{1 \over z^{n + 1}}
\sum_{x_{1} = -\infty}^{\infty}z^{\verts{x_{1}}}\cdots
\sum_{x_{k} = -\infty}^{\infty}z^{\verts{x_{k}}}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1^{-}}{1 \over z^{n + 1}}\,\pars{1 + 2\,{z \over 1 - z}}^{k}
\,{\dd z \over 2\pi\ic}
\\[3mm] = &\
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{k} \over z^{n + 1}}\,\pars{1 - z}^{-k}
\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{k} \over z^{n + 1}}\,\
\overbrace{\sum_{j = 0}^{\infty}{-k \choose j}\pars{-z}^{j}}^{\ds{\pars{1 - z}^{-k}}}\
\,{\dd z \over 2\pi\ic}
\\[3mm] = &\
\sum_{j = 0}^{\infty}{-k \choose j}\pars{-1}^{j}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{k} \over z^{n + 1 - j}}\,\
\,{\dd z \over 2\pi\ic}}^{\ds{{k \choose n - j}}} =
\sum_{j = n - k}^{n}{k + j - 1 \choose j}\pars{-1}^{j}\pars{-1}^{j}
{k \choose n - j}
\\[3mm] & =
\sum_{j = 0}^{k}{k + j + n - k - 1 \choose j + n - k}
{k \choose n - j - n + k} =
\color{#f00}{\sum_{j = 0}^{k}{j + n - 1 \choose k - 1}{k \choose  j}}
\end{align}