The number of possible outcomes adding up to $18$ in the experiment of tossing $3$ fair $6$-sided dice is $1$ - all dice showing $6$ or $(6,6,6).$
However, if I try to use generating functions, I am probably making a mistake somewhere...
$$G(x) = (x^1 + x^2 + x^3 + \cdots)^3=x^3(1 + x + x^2 + x^3+\cdots)^3=\frac{x^3}{(1-x)^3}$$
Now to extract the coefficient for $x^{18}$ of the series $(1 + x + x^2 + x^3+\cdots)^3$ the formula would be $\binom{18+3-1}{18}$. However, because there is an $x^3$ factored out, I would assume the correct formula would be:
$$\binom{18 - 3 +3 -1}{18-3}=\binom{17}{15}$$
which is different from the expected $1.$
What am I doing wrong?
The problem is that the infinite power series allows you to select partitions like $(9, 5, 4)$. You need to consider up to $x^6$ for each die, and not further. This finite restriction will make the numbers work out.
So:
$$G(x) = (x + x^{2} + \ldots + x^{6})^{3} = \left( \dfrac{1-x^{7}}{1-x} \right)^{3}$$
Edit: Recall that $(1-x^{7})^{3}$ can be expanded by the binomial theorem, and you are already familiar with the expansion of $\dfrac{1}{(1-x)^{3}}$. Now:
$$\left( \dfrac{1-x^{7}}{1-x} \right)^{3} = (1-x^{7})^{3} \cdot \dfrac{1}{(1-x)^{3}} = \left( \sum_{n=0}^{3} (-1)^{n}x^{7n} \right) \cdot \left( \sum_{n=0}^{\infty} \binom{n+3-1}{n} x^{n} \right)$$
Now by foiling out, deduce the coefficient of $x^{18}$.