Example 3 (Question from the 2nd "Hope Cup" Invitational Competition) If sets $M$ and $N$ each contain $m$ and $n$ elements, respectively, then the number of possible mappings from $M$ to $N$ is ().
A. $m+n$
B. $m \cdot n$
C. $m^{n}$
D. $n^{m}$
Solution: Choose D. Reason: Let $M=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, N=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, then $f\left(a_{i}\right)$ can be equal to any of $b_{1}$, $b_{2}, \cdots, b_{n}$, which gives $n$ possibilities, and $i=1,2, \cdots, m$, so the number of mappings sought is $n^{m}$.