How many non-negative integer solution is there for the equation $ x_1 + x_2 + x_3 + x_4 + 8*x_5 = 20 $? I am trying to figure out this problem and the way I think I might need to solve this is to account for the first 4 $x_i$ terms and then add on the possibilities for the $8x_5$ but I am not 100% sure that this is what I need to do. Any suggestions would be appreciated.

Partitioning for $x_5$ $$x_1 + x_2 + x_3 + x_4 + 8*x_5 = 20\\ \to 
8x_5 \leq 20 \to x_5\leq\frac{20}{8} \to x_5=0,1,2\\
\begin{cases}x_5=0 \to x_1 + x_2 + x_3 + x_4 + 8*0= 20\\\\x_5=1 \to x_1 + x_2 + x_3 + x_4 + 8*1= 20\\\\x_5=2 \to x_1 + x_2 + x_3 + x_4 + 8*2= 20\end{cases}\\$$
WE know non-negative integer solution is there for the equation $x_1+x_2+..=+x_k=n$ is $\color{red}{\left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)
}$ so 
$$ 
\begin{cases}x_5=0 \to x_1 + x_2 + x_3 + x_4 = 20&\left(\begin{array}{c}20+4-1\\ 4-1\end{array}\right)
\\\\x_5=1 \to x_1 + x_2 + x_3 + x_4 = 12&\left(\begin{array}{c}12+4-1\\ 4-1\end{array}\right)
\\\\x_5=2 \to x_1 + x_2 + x_3 + x_4 = 4 &\left(\begin{array}{c}4+4-1\\ 4-1\end{array}\right)
\end{cases}$$ 
final answer is :$$\left(\begin{array}{c}20+4-1\\ 4-1\end{array}\right)+\left(\begin{array}{c}12+4-1\\ 4-1\end{array}\right)+\left(\begin{array}{c}4+4-1\\ 4-1\end{array}\right)=\\\left(\begin{array}{c}21\\ 3\end{array}\right)+\left(\begin{array}{c}15\\ 3\end{array}\right)+\left(\begin{array}{c}7\\ 3\end{array}\right)$$
