Find coefficient of $x^r$ in $(x^5+x^6+x^7+...)^8$

Question

My work:

We can rewrite the generating function $(x^5+x^6+x^7+...)^8$ as $x^{40}(1+x+x^2+...)^8$

We are looking for $x^{r-40}$ coefficient in the new generating function $(1+x+x^2+...)^8$

We can rewrite $(1+x+x^2+...)^8$ as $(\cfrac{1}{1-x})^8$ = $(1-x)^{-8}$

So coefficient can be found by ${-8 \choose r-40} = {r - 33 \choose r-40 }$

so final answer is ${r - 33 \choose r-40 }$

Is this correct?

This is a concept question so I want to make sure I'm understanding the material properly.

Answer 1

There is a very close association between stars and bars and generating functions: see https://brilliant.org/wiki/integer-equations-star-and-bars/

Alternative approach

you want the # of non-negative integer solutions to

$x_1 + x_2 + x_3 + \cdots + x_8 = (r-40).$

This is immediately seen to be

$$\binom{[r-40] + [8-1]}{[8-1]}.$$