[Solution] . If $f(a)=f(b)=f(c)=f(d)=2$, it is clearly a mapping that fits the problem's requirements.

If $f(a), f(b), f(c), f(d)$ are not all 2, then at least one of them must be 1, and at most one of them can be 1 (note that $2^{3}=820$).
We discuss the following two scenarios:
(1) Among $f(a), f(b), f(c), f(d)$, there is one 1 and one 3 (two 2s), in this case, there are $P_{4}^{2}=4 \cdot 3=12$ mappings;
(2) Among $f(a), f(b), f(c), f(d)$, there is one 1 and two 3s (one 2), in this case, there are $P_{4}^{2}=4 \cdot 3=12$ mappings.

In total, there are $1+12+12=25$ mappings. Therefore, the answer is (C).
15$\cdot$18 Mapping $f:\{a, b, c, d\} \rightarrow\{1,2,3\}$. If $10<f(a) \cdot f(b)$. $f(c) \cdot f(d)<20$, the number of such mappings is
(A) 23.
(B) 24.
(C) 25.
(D) 26.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)