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.

Rewrite as $x^{40}((1+x^5)+x^7)^{20}$, then the k-th term of the binomial expansion (ignoring the $x^{40}$) is $\binom{20}kx^{7(20-k)}\sum_{i=0}^k \binom kix^{5i}$.
We now actually want the exponent to be 17. Check $i=0, 1, 2, 3$ to see the only possibility is $i=2$, which gives $k=19$.
Hence the coefficient is $\binom{20}{19}\binom{19}2=3420.$
