Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?

My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.

Can you please explain?


It's basically the number of ways you can write $12$ as a sum of three integers all greater than or equal to $3$, where order matters.  So $12=3+3+6$, $12=3+4+5$, $12=4+4+4$.  The first can be rearranged three ways, the second can be rearranged six ways and the last can only be arranged one way.  So yes, I believe the answer is $10$.
