3. 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 $B$ makes every element in $B$ have a preimage, and $f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant$ $f\left(a_{100}\right)$, then the number of such mappings is ( ).
(A) $\mathrm{C}_{100}^{50}$
(B) $\mathrm{C}_{90}^{50}$
(C) $\mathrm{C}_{1 \infty}^{49}$
(D) $\mathrm{C}_{99}^{49}$
(2002, National High School Mathematics Competition)
(Tip: Let $b_{1}<b_{2}<\cdots<b_{50}$, 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 problem's requirements, and each such grouping corresponds one-to-one to a mapping that satisfies the conditions. Therefore, the number of mappings $f$ that satisfy the problem's requirements is equal to the number of ways to divide $A$ into 50 groups in order. The number of ways to divide $A$ is $\mathrm{C}_{99}^{49}$, so there are $\mathrm{C}_{99}^{99}$ such mappings. Hence, the answer is (D).)