What's the coefficent of $x^{20}$ in $(x^3+x^4+x^5+...)^5$? Only hint is needed. I know about Binomial/Multinomial expansion but I got stuck on this series; it doesn't look like anything I've solved before. I already searched for any hint/formula and couldn't find one, any help is appreciated.

Can be written as 
$x^{15}(1-x)^{-5}$
Use binomial expansion for $(1-x)^{-5}$
That is
$$x^{15}(1+5x+ \frac{5 \cdot 6}{2!}x^2 + \frac{5 \cdot 6 \cdot 7}{3!}x^3 + \frac{5 \cdot 6 \cdot 7 \cdot 8}{4!}x^4 + \frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}x^5 + \cdots)$$
Coefficient of $x^{20}$ is 
$x^{15}×$ coefficient of $x^5$ .
Which is $$\frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}$$ 
