I am having problems understanding the logic in solving the following question:
Find the generating function for the number of integer solutions to the equation : $c_1 + c_2 + c_3 +c_4 = 20$
where,
$ \\c_1 \ge -3 \\c_2 \ge -3 \\ -5\le c_3 \le 5 \\ c_4 \ge 0$
Solution: The coefficient of $x^{31}$ in the generating function $f(x)=(1+x+x^2 + x^3 +x^4+\cdots)^3(1+x+x^2+\cdots+x^{10})$
I believe I have an understanding on the first term inside of the generating funciton, which was extracted from the condition $c_1, c_2, c_4$ due to the conditions $c_4 \ge 0$ and $c_1 c_2$ having no upper limit resulting in the trailing periods ($+\cdots$).
What I do not understand is what is happening to $c_2$? How does it form the the polynomial $(1+x+x^2+\cdots+x^{10})$.
What effect does negative integer restrictions have in forming generating functions?
Thank you for all your help, please let me know if any step is incorrect.
Here is how I would approach this. introduce new variables $k_1 = c_1+3 \ge 0, k_2 = c_2+3, k_3 = c_3+5$ and your equation becomes
$$20 = \sum c_i = \sum k_i - 11$$ or equivalently, $$\sum k_i = 31$$ which you can solve in the usual way since all $k_i \ge 0$...