21311 ㅊ Let the three-digit number $n=\overline{a b c}$, if $a, b, c$ can form an isosceles (including equilateral) triangle, find the number of such three-digit numbers $n$.


To analyze $a$, $b$, and $c$ to form the sides of a triangle, it is clear that none of them can be 0, i.e., $a, b, c \in \{1, 2, \cdots, 9\}$.
(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, $n_{1} = \mathrm{C}_{9}^{1} = 9$.
(2) If they form an isosceles (non-equilateral) triangle, let the number of such three-digit numbers be $n_{2}$. Since only 2 different digits are used in the three-digit number, let these digits be $a$ and $b$. Noting that the legs and base of the triangle can be interchanged, the number of digit pairs $(a, b)$ is $2 \mathrm{C}_{9}^{2}$. However, when the larger number is the base, if $a > b$, it must satisfy $b < a < 2b$. In this case, the digit pairs that cannot form a triangle are as follows, totaling 20 cases:

At the same time, for each digit pair $(a, b)$, the two digits can be filled into the three positions in $\mathrm{C}_{3}^{2}$ ways. Therefore, $n_{2} = \mathrm{C}_{3}^{2} \left(2 \mathrm{C}_{9}^{2} - 20\right) = 6 \left(\mathrm{C}_{9}^{2} - 10\right) = 156$.
In summary, $n = n_{1} + n_{2} = 165$.