How many integer solutions are there of the equation $|x_{1}|+|x_{2}|+\cdots +|x_{k}|=n$? 
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 $\binom{n+k-1}{k-1}$  ) 
 A: It is a simple variation on stars and bars. The number of solutions of
$$ x_1+x_2+\ldots+x_k = n $$
with $x_i\in\mathbb{N}^+$ is given by the coefficient of $x^n$ in $\left(\frac{x}{1-x}\right)^k$, i.e. by $\binom{n-1}{k-1}$, so the number of solutions of
$$\left|x_1\right|+\ldots +\left|x_k\right| = n$$
with $x_i\in\mathbb{Z}\setminus\{0\}$ is given by $2^k\binom{n-1}{k-1}$. If we take the number of zero variables as a parameter, we get that the number of solutions of
$$ \left|x_1\right|+\ldots +\left|x_k\right| = n $$
with $x_i\in\mathbb{Z}$ is given by:

$$ \sum_{h=0}^{k}\binom{k}{k-h}\binom{n-1}{h-1}2^h. $$

The connection with the GFs found by Felix and Semiclassical lies here:
$$\begin{eqnarray*}\sum_{h=0}^{k}\binom{k}{h}2^h [x^{n-h}]\left(\frac{x}{1-x}\right)^h &=& \sum_{h=0}^{k}\binom{k}{h}[x^n]\left(\frac{2x}{1-x}\right)^h\\&=&[x^n]\left(1+\frac{2x}{1-x}\right)^k\\&=&[x^n]\left(\frac{1+x}{1-x}\right)^k.\end{eqnarray*}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\delta_{\ell\ell'}}$ 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}
A: A generating-function solution is straightforward. For each $|x_j|$, every positive $n\in\mathbb{Z}$ occurs as both $x_j=n$ and $x_j=-n$; on the other hand, $|x_j|=0$ requires $x_j=0$. This sequence is generated by $$1+2t+2t^2+\cdots =(1+t)(1-t)^{-1}.$$ (This can be checked by multiplying both sides by $1-t$.) 
Counting the solutions to $\sum_{j=1}^k |x_j|=n$ then amounts to finding the  coefficient of $t^n$ in $\left(\dfrac{1+t}{1-t}\right)^k.$ But these can be found as the Cauchy product of the coefficients of $(1+t)^k$ and $(1-t)^{-k}$, each of which can be found by the binomial theorem. This should yield the same answer as obtained by Jack D'Aurizio; I leave the remaining details to the interested reader.
