Find number of solutions for integer equation: way to simplify calculations?

Question

Find number of solutions of the equation:

$$ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20, \quad 0 \leq x_i \leq 8, \quad x_i \in \Bbb Z $$

The answer $(27237)$ can be found with generative polynomial: $$ (1 + n + n^2 + \cdots + n^8)^6 = {}\cdots + 27237n^{20} + \cdots {} $$

Is there any way to simplify calculations? Expanding above expression by hand (even discarding higher powers of $n$ whenever possible) is rather long.

It's possible to calculate by hand number of solutions with nines, tens, eleven, twelve with combinations with repetitions and get the answer from there, but that seems like less elegant/generalized.

Maybe what I'm trying to ask is: is there a way to get a specific coefficient of a polynomial without fully expanding all the lower powers.

Answer 1

Here is a way that does not seem to be so lengthy. Denoting with $[n^k]$ the coefficient of $n^k$ of a series, we obtain \begin{align*} \color{blue}{[n^{20}]}&\left(1+n+n^2+\cdots+n^8\right)^{6}\\ &=[n^{20}]\left(\frac{1-n^9}{1-n}\right)^6\\ &=[n^{20}]\left(1-\binom{6}{1}n^9+\binom{6}{2}n^{18}\right)\frac{1}{(1-n)^6}\\ &=\left([n^{20}]-6[n^{11}]+15[n^{2}]\right)\sum_{k=0}^{\infty}\binom{k+5}{k}n^k\\ &=\binom{25}{20}-6\binom{16}{11}+15\binom{7}{2}\\ &=53\,130-6\cdot4\,368+15\cdot21\\ &\,\,\color{blue}{=27\,237} \end{align*} according to the claim.

Answer 2

By multinomial theorem, direct calculation is quite hard. A way to simplify it is as follow.

Since $9$ is not a prime $f(x)=1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8$ could be reducible maybe. So is in fact and we have $f(x)=(1+x+x^2)(1+x^3+x^6)=A\times B$ so $[f(x)]^6=A^6\times B^6$. It follows

$$A^3=(1+x+x^2)^3=x^6+3x^5+6x^4+7x^3+6x^2+3x+1$$ $$B^3=(1+x^3+x^6)^3=x^{18}+3x^{15}+6x^{12}+7x^9+6x^6+3x^3+1$$ $$A^6=x^{12}+6(x^{11}+x)+21(x^{10}+x^2)+50(x^9+x^3)+90(x^8+x^4)+126(x^7+x^5)+141x^6+1$$ $$B^6=1+6x^3+21x^6+50x^9+90x^{12}+126x^{15}+141x^{18}+X$$ where $X=x^{21}(126+90x^3+50x^6+21x^9+6x^{12})$ so we have to calculate the coefficient of $x^{20}$ in the product (without $X$) $$(1+6x^3+21x^6+50x^9+90x^{12}+126x^{15}+141x^{18})\times(x^{12}+6(x^{11}+x)+21(x^{10}+x^2)+50(x^9+x^3)+90(x^8+x^4)+126(x^7+x^5)+141x^6+1)$$ which is easy. We have $$141\times21+126^2+90^2+50\times6=\boxed{27237}$$