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Use exponential generating functions to determine the number $a_n$ of ordered choices of $n$ balls such that there are $2$ or $4$ red balls, an even number of green balls, and an arbitrary number of blue balls.

If we denote red with $r$, green with $g$, and blue with $b$. I guess we will have $r^2 + r^4$ for red balls, $1 + b + b^2 + \ldots = \frac{b}{1-b}$ for blue balls. But how can I express the green balls, and how can I then solve the problem using exponential generating functions?

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Exponential generating functions are useful for this sort of problem because we can find the answer by multiplying the EGF for each of the types of balls: $$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$ This expands to $$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. $$ Simplification yields $$\frac{x^2}{2!}+3\frac{x^3}{3!}+13\frac{x^4}{4!} + \sum_{n=5}^\infty \frac1{768} 2^nn(n-1)(n^2-5n+54)\frac{x^n}{n!}. $$ Hence $$ a_n =\begin{cases} 1,& n=2\\ 3,& n=2\\ 13,& n=4\\ \frac1{768} 2^nn(n-1)(n^2-5n+54),& n\geqslant 5. \end{cases} $$

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