Given two sets of real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\}$, if the mapping $f$ from $A$ to $B$ ensures that each element in $B$ has a preimage, and
$$
f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant f\left(a_{100}\right)
$$

Find the number of such mappings.
Parse: Not setting $b_{1}<b_{2}<\cdots<b_{50}$, divide the elements $a_{1}, a_{2}, \cdots, a_{100}$ of $A$ in order into 50 non-empty groups. Define the mapping $f: A \rightarrow B$, such that the image of the elements in the $i$-th group under $f$ is all $b_{i}(i=1$, $2, \cdots, 50$). It is easy to see that such an $f$ meets the requirements of the problem, and all such groupings correspond one-to-one with the mappings that satisfy the conditions. Therefore, the number of mappings $f$ that meet the requirements is equal to the number of ways to divide $A$ into 50 groups in order of subscripts, and the number of ways to divide $A$ is $\mathrm{C}_{99}^{49}$. Thus, there are $\mathrm{C}_{99}^{49}$ such mappings.