I need to find the number of solutions to the following equation
$ x_1+x_2+x_3+x_4 = n $ where $ x_1 = 2k, \space x_2 \in \{0,1,2\}, \space x_3=3q, \space x_4 \in \{0,1\}$
Using generating functions. The biggest problem I'm having with approaching this problem is the fact that the equation is being set equal to $n$ and not a specific number.
Thanks!
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The solution is given by \begin{align} &\bracks{z^{n}}\braces{\pars{\sum_{x_{1} = 0}^{\infty}z^{2x_{1}}} \pars{z^{0} + z^{1} + z^{2}}\pars{\sum_{x_{3} = 0}^{\infty}z^{3x_{3}}} \pars{z^{0}+z^{1}}} \\[5mm] = &\ \bracks{z^{n}}\braces{{1 + z \over 1 - z^{2}}\,{1 + z + z^{2} \over 1 - z^{3}}} = \bracks{z^{n}}{1 \over \pars{1 - z}^{2}} = \bracks{z^{n}}\sum_{k = 0}^{\infty}{-2 \choose k}\pars{-z}^{k} \\[5mm] = &\ \bracks{z^{n}}\sum_{k = 0}^{\infty}{k + 1 \choose k}\pars{-1}^{k}\pars{-z}^{k} = \bracks{z^{n}}\sum_{k = 0}^{\infty}\pars{k + 1}z^{k} = \bbx{\ds{n + 1}} \end{align}