Example 4.8 Find the number of positive integer solutions to the equation $x_{1}+2 x_{2}+4 x_{3}=21$.

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Solution: Let the required value be $N$, then $N$ is
$$
\begin{aligned}
A(t)= & \left(t+t^{3}+t^{5}+\cdots\right) \cdot\left(t^{2}+t^{4}+t^{6}+\cdots\right) \cdot \\
& \left(t^{4}+t^{8}+t^{12}+\cdots\right)
\end{aligned}
$$

the coefficient of $t^{21}$ in the expansion, and
$$
\begin{aligned}
A(t) & =t\left(1-t^{2}\right)^{-1} \cdot t^{2}\left(1-t^{2}\right)^{-1} \cdot t^{4}\left(1-t^{4}\right)^{-1} \\
& =t^{7}\left(1-t^{2}\right)^{-2}\left(1-t^{4}\right)^{-1} \\
& =t^{7}\left(1+t^{2}\right)^{2}\left(1-t^{4}\right)^{-3} \\
& =\left(t^{7}+2 t^{9}+t^{11}\right) \sum_{k=0}^{\infty}\binom{k+2}{2} t^{4 k} .
\end{aligned}
$$

Since $7+4 k=21$ has no integer solution, $9+4 k=21$ has the solution $k=3$, and $11+4 k=21$ has no integer solution, therefore
$$
N=2 \cdot\binom{3+2}{2}=20 \text {. }
$$