Generating functions to solve number of integer solution problem If I have $x_1 + x_2 + x_3 =10$ with $1\leq x_1 \leq 5, \; 2 \leq x_2 \leq 6, \;3 \leq x_3 \leq 9$
I know that I compute $(t^1+\dots + t^5)(t^2 +\dots + t^6)(t^3+\dots +t^9)$ and look at the coefficient of $t^{10}$ to find number of integer solutions.
But below I have $3a_2,5a_3$
Say I have $a_1+3a_2+5a_3=33$ with $1\leq a_1 \leq 11,\;3\leq a_2 \leq 18,\; 5 \leq a_3 \leq 13$.
Do I simply change it to $9 \leq 3a_2 \leq 54$ etc
and compute $(t^1+t^2+\dots +t^{10}+ t^{11})(t^9+t^{12}+\dots +t^{51}+t^{54})(t^{25}+t^{30}+\dots +t^{60}+t^{65})$ and look at coefficient $33$?

Meaning do I count the exponents up in the increased rate?


You are looking for ($\dotsb$ at the end of a factor covers terms that don't affect the result):
$\begin{align}
[z^{10}] (z + \dotsb + z^5) &(z^2 + \dotsb + z^6) (z^3 + \dotsb + z^9) \\
  &= [z^{10 - 1 - 2 - 3}] (1 + \dotsb + z^4) (1 + \dotsb + z^4) (1 + \dotsb + z^6) \\
  &= [z^6] \frac{(1 - z^5)^2 (1 - z^7)}{(1 - z)^3} \\
  &= [z^6] (1 - 2 z^5 - \dotsb) \sum_{k \ge 0} (-1)^k \binom{-3}{k} z^k \\
  &= [z^6] (1 - 2 z^5) \sum_{k \ge 0} \binom{k + 3 - 1}{3 - 1} z^k \\
  &= \binom{6 + 2}{2} - 2 \binom{6 - 5 + 2}{2} \\
  &= 22
\end{align}$
