Example 8 Let the three-digit number $n=\overline{a b c}$, if the lengths of the sides of a triangle can be formed with $a, b, c$ as an isosceles (including equilateral) triangle, then the number of such three-digit numbers $n$ is $\qquad$.
(2004 National High School League Question, originally a multiple-choice question)
(1) If they form an equilateral triangle, then the number of such three-digit numbers is $n_{1}=C_{9}^{1}=9$.
(2) If they form an isosceles (non-equilateral triangle), let the number of such three-digit numbers be $n_{2}$. Since there are only 2 different digits in the three-digit number, denoted as $a, b$, note that the legs and base of the triangle can be swapped, so the number of digit pairs $(a, b)$ is $2 C_{9}^{2}$. However, when the larger number is the base, assuming $a>b$, it must satisfy $b<a<2b$. At this time, the digit pairs that cannot form a triangle are
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$b$ & 4,3 & 4,3 & 3,2 & 3,2 & 2,1 & 2,1 & 1 & 1 & \\
\hline
\end{tabular}

There are a total of 20 such cases.
At the same time, for each digit pair $(a, b)$, the two digits can be filled into three positions in $C_{3}^{2}$ ways.
Therefore, $n_{2}=C_{3}^{2}\left(2 C_{9}^{2}-20\right)=156$. Hence, the number of three-digit numbers that satisfy the conditions is $n_{1}+n_{2}=165$.