Determine how many ways Brian, Katie, and Charlie can split a $\$50$ dinner bill such that Brian and Katie each pay an odd number of dollars and Charlie pays at least $\$5$.
How can I use generating functions to solve this problem? Or should I use a counting approach?
On
The generating function for the number of ways that the three can pay a bill of $n$ dollars is $$ \begin{align} &(x+x^3+x^5+\dots)(x+x^3+x^5+\dots)(x^5+x^6+x^7+\dots)\\[3pt] &=x^7\frac1{(1-x^2)^2}\frac1{1-x}\\ &=x^7\frac1{(1-x)^3}\frac1{(1+x)^2}\\ &=x^7\left[\frac14\frac1{(1-x)^3}+\frac14\frac1{(1-x)^2}+\frac3{16}\frac1{1-x}+\frac18\frac1{(1+x)^2}+\frac3{16}\frac1{1+x}\right]\\ &=\sum_{k=0}^\infty\small\left[\frac14(-1)^k\binom{-3}{k}+\frac14(-1)^k\binom{-2}{k}+\frac3{16}(-1)^k\binom{-1}{k}+\frac18\binom{-2}{k}+\frac3{16}\binom{-1}{k}\right]x^{k+7}\\ &=\sum_{k=0}^\infty\small\left[\frac14\binom{k+2}{2}+\frac14\binom{k+1}{1}+\frac3{16}\binom{k}{0}+\frac18(-1)^k\binom{k+1}{1}+\frac3{16}(-1)^k\binom{k}{0}\right]x^{k+7}\\ &=\sum_{k=0}^\infty\frac1{16}\left[(2k^2+10k+11)+(-1)^k(2k+5)\vphantom{\frac1{16}}\right]x^{k+7}\\ &=\sum_{k=0}^\infty\left[\binom{\lfloor k/2\rfloor}{2}+2\binom{\lfloor k/2\rfloor}{1}+\binom{\lfloor k/2\rfloor}{0}\right]x^{k+7}\\ &=\sum_{k=0}^\infty\binom{\lfloor k/2\rfloor+2}{2}x^{k+7}\tag{1} \end{align} $$ Plugging $k=43$ into $(1)$ yields the coefficient of $x^{50}$ to be $253$.
Take $\$6$ off Charlie and a dollar off the other two. Now lay out the remaining $\$42$ in $21$ $\$2$ chits and divide those. By stars-and-bars, there are
$${21+2 \choose 2} = 253\text{ options}$$
I'll leave the generating functions to someone else...