<s>[INST] Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)^{20}$ and then working with binomial expansion.
 [/INST] There is a thing called multinomial theorem which is a slight generalization of the binomial theorem.
First of all, note $(x^2+x^7+x^9)^{20}= x^{40}(1+x^5+x^7)^{20}.$ So, basically, now will find the coefficients of $x^{17}.$ Can you continue?
Note that $(1)^{17} (x^5)^2(x^7)^1$ gives us $x^{17}.$ The coffocient of  $x^{17}$ is $\dfrac{20!}{17! 2! 1!}$. 
The idea of binomial/multinomial theorem is purely combinatorial. 
Suppose we have $(x+y+z)^{20}.$ Consider the term $x^{10}y^{6}z^{4}.$ Note that the coefficient of this term is $\dfrac{20!}{10! 6! 4!}$ The combinatorial interpretation is that you can create/form the $x^{10}y^{6}z^{4}$ $\dfrac{20!}{10! 6! 4!}$ ways. This should you give you some idea to think more about the problem. I hope it helps.
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