Example 16 (2002 National High School Competition Question) Given two sets of real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \cdots\right.$, $\left.b_{50}\right\}$. If the mapping $f$ from $A$ to $B$ makes each element in $B$ have a preimage, and $f\left(a_{1}\right) \leqslant \cdots \leqslant f\left(a_{2}\right) \leqslant f\left(a_{100}\right)$, then the number of such mappings is ( ) .
B. $C_{99}^{18}$
C. $C_{100}^{99}$
D. $C_{99}^{49}$
Solution: Choose D. Reason: Suppose $b_{1}<b_{2}<\cdots<b_{50}$, and divide the elements $a_{1}, a_{2}, \cdots, a_{100}$ of $A$ into 50 non-empty groups in order. Define the mapping $f: A \rightarrow B$ such that the image of the elements in the $i$-th group under $f$ is $b_{i}(i=1,2, \cdots, 50)$. It is easy to see that such an $f$ satisfies the requirements of the problem, and each such grouping corresponds one-to-one to a mapping that satisfies the conditions. Therefore, the number of mappings $f$ that satisfy the requirements of the problem is equal to the number of ways to divide $A$ into 50 groups in order, and the number of ways to divide $A$ is $C_{99}^{49}$, so there are $C_{99}^{49}$ such mappings.