We might as well suppose $b_1 < b_2 < \dots < b_{50}$, and divide elements $a_1, a_2, \dots, a_{100}$ in $A$ into 50 nonempty groups according to their order. Define a mapping $f: A \to B$, so that the images of all the elements in the $i$-th group are $b_i$ ($i = 1, 2, \dots, 50$) under the mapping. Obviously, $f$ satisfies the requirements given in the problem. Furthermore, there is a one-to-one correspondence between all groups so divided and the mappings satisfying the condition. So the number of mappings $f$ satisfying the requirements is equal to the number of ways dividing $A$ into 50 groups according to the order of the subscripts. The number of ways dividing $A$ is $C_{99}^{49}$.

Then there are, in all, $C_{99}^{49}$ such mappings. Answer: D.