Analysis: Since the permutation of $n$ elements is an arithmetic expression, the last symbol from left to right must be a digit, and the $(n-1)$-th position has two possibilities: it is either a digit, in which case the first $(n-1)$ positions already form an arithmetic expression. If the $(n-1)$-th position is an operator, then the first $(n-2)$ positions form an arithmetic expression.
Solution: Let $f_{n}$ represent the number of ways to arrange $n$ elements into an arithmetic expression, then we have
$$
\left\{\begin{array}{l}
f_{n}=9 f_{n-1}+36 f_{n-2}, \\
f_{1}=9, f_{2}=99 .
\end{array}\right.
$$
$f_{2}=99$ refers to the 81 numbers from $11 \cdots 19, 21 \cdots 29, \cdots 91 \cdots 99$, and the 18 arithmetic expressions formed by $\pm 1$, $\cdots, \pm 9$.
The characteristic equation is $\lambda^{2}-9 \lambda-36=0$,
$$
\lambda=12,-3, \quad f_{n}=k_{1} \times 12^{n}+k_{2}(-3)^{n},
$$

From $f_{1}=9, f_{2}=99$, we can deduce that $f_{0}=\frac{1}{2}$.
Using $f_{0}=\frac{1}{2}, f_{1}=9$, we find $k_{1}=\frac{7}{10}, k_{2}=-\frac{1}{5}$,
$$
\therefore \quad f_{n}=\frac{7}{10} \times(12)^{n}-\frac{1}{5} \times(-3)^{n} .
$$
Example 2.4.5 Nine digits (1 to 9) and 4 arithmetic operators $(+,-, \times, \div)$ make up 13 elements. Find the number of arithmetic expressions that can be formed by arranging $n$ of these elements.