2B. In a certain populated place, all telephone numbers consist of six digits arranged in strictly ascending or strictly descending order, with the first digit in the number not being 0. What is the maximum number of telephone numbers that can exist in this place?
Solution. Let's set $n=n_{1}+n_{2}$, where $n_{1}$ is the number of telephone numbers with digits in strictly increasing order, and $n_{2}$ is the number of telephone numbers with digits in strictly decreasing order. Each number whose digits are in strictly decreasing order is obtained by selecting six digits out of a total of 10, which are then arranged from the largest to the smallest. Therefore, such numbers can total $n_{2}=\binom{10}{6}=210$. Similarly, each number whose digits are in strictly increasing order is obtained by selecting the digits and then arranging them from the smallest to the largest. Here, we have 9 digits available, as numbers cannot start with 0, which would be the smallest digit in any possible combination that includes it. Thus, $n_{1}=\left({ }_{6}^{9}\right)=84$. Therefore, the maximum number of telephone numbers that can exist in this place is

$$
n=84+210=294
$$

NOTE. Prove that

$$
\frac{1}{2021}<\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{2019}{2020}<\sqrt{\frac{1}{2021}}
$$