$10 \cdot 2$ In the numbers $100, 101, \cdots, 999$, the number of numbers whose digits are in strictly increasing or strictly decreasing order is
(A) 120.
(B) 168.
(C) 204.
(D) 216.
(E) 240.
(41st American High School Mathematics Examination, 1990)
[Solution] First, consider the case where the digits are strictly increasing:
Table for numbers starting with 1
\begin{tabular}{r|c|c|c|c|c|c|c}
\hline Interval & $123-129$ & $134 \sim 139$ & $145-149$ & $156 \sim 159$ & $167-169$ & $178-179$ & 189 \\
\hline Number & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline
\end{tabular}

Similarly, for numbers starting with 2, there are $6,5,4,3,2,1$ each;
For numbers starting with 3, there are $5,4,3,2,1$ each;
For numbers starting with 4, there are $4,3,2,1$ each;
For numbers starting with 5, there are $3,2,1$ each;
For numbers starting with 6, there are 2,1 each;
For numbers starting with 7, there is only 1.
In total, there are $1+3+6+10+15+21+28=84$;
Due to symmetry, numbers in strictly decreasing order (excluding those ending in 0) also have 84.

Consider $980,970, \cdots, 910 ; 870,860, \cdots, 810 ; \cdots ; 210$. There are $8,7,6$, $5,4,3,2,1$ each, totaling 36.
Thus, the total number of numbers in the problem is $84 \cdot 2+36=204$. Therefore, the answer is (C).