$15 \cdot 17$ If sets $M, N$ each contain $m, n$ elements, then the number of mappings that can be established from $M$ to $N$ is
(A) $m+n$.
(B) $m \cdot n$.
(C) $m^{n}$.
(D) $n^{m}$.
(2nd "Hope Cup" National Mathematics Invitational, 1991)
[Solution] For each element in $M$, $n$ different mappings can be established from $M$ to $N$:
That is, if $M=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}, N=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, then
$$
a_{k} \rightarrow b_{l} \quad(k=1,2, \cdots, m ; \quad l=1,2, \cdots, n)
$$

Therefore, the number of mappings from $M$ to $N$ is
$$
\underbrace{n \cdot n \cdot \cdots \cdot n}_{m \uparrow}=n^{m} \text {. }
$$

Hence, the correct choice is $(D)$.