Example 3 Let the set $A=\{1,2, \cdots, 10\},$ and the mapping $f$ from $A$ to $A$ satisfies the following two conditions:
(1) For any $x \in A, f_{30}(x)=x$;
(2) For each $k \in \mathbb{Z}^{+}, 1 \leqslant k \leqslant 29$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.
Find the total number of such mappings.
(1992, Japan Mathematical Olympiad Preliminary)
Solution: Notice that $10=5+3+2,30=5 \times 3 \times 2$. This suggests dividing $A$ into three disjoint subsets
$$
A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\} \cup\left\{b_{1}, b_{2}, b_{3}\right\}
$$
$\cup\left\{c_{1}, c_{2}\right\}$.
Since $f$ satisfies conditions (1) and (2), $f$ is a one-to-one mapping from $A$ to $A$, and from the proof process of Proposition 1, we know that there exists a mapping cycle in $A$. Therefore, define the mapping
$$
\begin{array}{l}
f: f\left(a_{1}\right)=a_{2}, f\left(a_{2}\right)=a_{3}, f\left(a_{3}\right)=a_{4}, \\
f\left(a_{4}\right)=a_{5}, f\left(a_{5}\right)=a_{1} ; f\left(b_{1}\right)=b_{2}, f\left(b_{2}\right) \\
=b_{3}, f\left(b_{3}\right)=b_{1} ; f\left(c_{1}\right)=c_{2}, f\left(c_{2}\right)=c_{1} .
\end{array}
$$

Since 30 is the least common multiple of $5,3,2$, by Proposition 1 and Proposition 2, we know that $f$ is the unique class of mappings that satisfy conditions (1) and (2) of the problem.

Therefore, the total number of $f$ is equivalent to selecting 5 elements from 10, then selecting 3 from the remaining 5, and finally selecting the last 2, and arranging them in circular permutations. It is equal to
$$
\left(C_{10}^{5} \cdot 4!\right)\left(C_{5}^{3} \cdot 2!\right)\left(C_{2}^{2} \cdot 1!\right)=120960
$$