Example 2-32 7 distinct balls are placed into 4 labeled boxes, with the requirement that boxes 1 and 2 must contain an even number of balls, and box 3 must contain an odd number of balls. Find the number of ways to do this.
This problem is equivalent to taking 7 elements from the set $\{1,2,3,4\}$ with repetition allowed, forming a permutation. Additionally, it is required that boxes 1 and 2 contain an even number of balls, box 3 contains an odd number of balls, and box 4 has no special requirements. For example, 3124214 is a valid permutation, which can be visualized as follows with 4 boxes:
$2,6 \quad 3,5 \quad 4,7$

This means the ball marked 1 is in box 3, the ball marked 2 is in box 1, the ball marked 3 is in box 2, and so on. Thus, 3124214 represents the sequence of placing balls marked $1,2,3,4,5,6,7$ into the boxes. According to the requirements, let $a_{n}$ be the number of ways to distribute $n$ balls. The exponential generating function for the sequence $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ is
$$
\begin{array}{l}
\begin{array}{l} 
G_{\mathrm{e}}(x)=\left(1+\frac{1}{2!} x^{2}+\frac{1}{4!} x^{4}+\cdots\right)^{2} \\
\text { corresponding to boxes (1) and (2) } \\
=\left(\frac{\mathrm{e}^{-x}+\mathrm{e}^{-x}}{2}\right)^{2}\left(\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right) \mathrm{e}^{x}
\end{array} \\
=\frac{1}{8}\left(\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}+2\right)\left(\mathrm{e}^{x}-\mathrm{e}^{-x}\right) \mathrm{e}^{x} \\
=\frac{1}{8}\left(\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}+2\right)\left(\mathrm{e}^{2 x}-1\right) \\
=\frac{1}{8}\left(\mathrm{e}^{4 x}-1+\mathrm{e}^{2 x}-\mathrm{e}^{-2 x}\right) \\
=\frac{1}{8}\left\{-1+\sum_{h=1}^{\infty}\left[4^{h}+2^{h}-(-2)^{h}\right] \frac{x^{h}}{h!}\right\} \\
a_{n}=\frac{1}{8}\left[4^{n}+2^{n}-(-2)^{n}\right], \quad n \geqslant 1 \\
\end{array}
$$