3. Among all six-digit natural numbers, the digits of which are arranged in ascending order (from left to right), numbers containing the digit 1 and not containing this digit are considered. Which numbers are more and by how many?
Solution. First, let's calculate how many six-digit natural numbers there are in total, with their digits arranged in ascending order. For this, we will write down all the digits from 1 to 9 in a row. To get six-digit numbers of the considered type, we need to strike out any three digits. Thus, the number of six-digit natural numbers, with their digits arranged in ascending order, is $C_{9}^{3}=84$.

Now let's calculate how many of these numbers contain 1. For this, we fix the digit 1 in our row of digits and strike out any three digits from the remaining 8. We get that the number of numbers containing 1 is $C_{8}^{3}=56$. As a result, we find that the numbers containing 1 are 28 more than the numbers that do not contain 1.

Answer: the numbers containing 1 are 28 more.