<s>[INST] How can I find the coefficient of a term in this generating function by using the "old" method? I'm trying to verify my answer to this problem by going back and solving the same problem using methods I've used before learning about generating functions, but I'm not quite sure how to do it in this case:

Find the coefficient of $x^{47}$ in $(x^{10}+x^{11}+\cdots+x^{25})(x+x^{2}+\cdots+x^{15})(x^{20}+\cdots+x^{45})$.

I'm looking at this problem as if it were saying

Given 47 identical objects and 3 distinct boxes, how many ways can the objects be placed into the boxes given that the first box must hold at least 10 and at most 25 objects, the second box must hold at least one and at most 15 objects, and the third box must hold at least 20 and at most 45 objects.

This can be translated to solving the equation
\begin{align}
e_1+e_2+e_3=47\\10\leq e_1\leq 25\\1\leq e_2\leq 15,\\20\leq e_3\leq 45\end{align}
If there were no upper limits on the capacity of the boxes, I'd say that the solution is ${47-10-1-20+2\choose 2}={18\choose 2}$. I get this by placing 2 "walls" between the three groups to separate the one group of 18 into 3, but I don't know how to do this considering there are upper bounds. The solution I got through solving directly from the generating function, the coefficient of $x^{47}$ is
$$(-1){2\choose 2}+(-1){3\choose 2}+(1){18\choose 2}$$
so where do those extra binomial coefficients come from?
 [/INST] You want (freely simplifying by getting rid of terms that can't affect the result):
\begin{align}
[x^{47}](x^{10}+x^{11}&+\cdots+x^{25})(x+x^{2}+\cdots+x^{15})(x^{20}+\cdots+x^{45}) \\
  &= [x^{47}] \; x^{10} (1 + x + \cdots + x^{10})
                \cdot x (1 + x + \cdots + x^{14})
                \cdot x^{20} (1 + x + \cdots + x^{25})\\
  &= [x^{16}] \frac{1 - x^{11}}{1 - z} 
                \cdot \frac{1 - x^{15}}{1 - z}
                \cdot \frac{1}{1 - z} \\
  &= [x^{16}] (1 - x^{11} - x^{15} + x^{26}) (1 - x)^{-3} \\
  &= [x^{16}](1 - x)^{-3} - [x^{5}] (1 - x)^{-3} - [x](1 - x)^{-3} \\
  &= (-1)^{16} \binom{-3}{16} - (-1)^5 \binom{-3}{5} - (-1) \binom{-3}{1} \\
  &= \binom{16 + 3 - 1}{3 - 1} - \binom{5 + 3 - 1}{3 - 1} - \binom{1 + 3 - 1}{3 - 1} \\
  &= 129
\end{align}
</s>