G2.3 設 $a_{1} 、 a_{2} 、 a_{3} 、 a_{4} 、 a_{5} 、 a_{6}$ 為非負整數, 並霂足 $\left\{\begin{array}{c}a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}=26 \\ a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=5\end{array}\right.$ 。若 $c$ 為方程系統的解的數量, 求 $c$ 的值。
Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be non-negative integers and satisfy
$$
\left\{\begin{array}{rl}
a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6} & =26 \\
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6} & =5
\end{array} .\right.
$$

If $c$ is the number of solutions to the system of equations, determine the value of $c$.
Let $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ be a solution, then $0 \leq a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \leq 5$
The solutions are:
$$
\begin{array}{l}
(0,1,0,0,0,4),(0,0,1,0,1,3),(0,0,0,2,0,3),(0,0,0,1,2,2),(0,0,0,0,4,1) \\
c=5
\end{array}
$$