11.2. Find the number of five-digit numbers containing two digits, one of which is divisible by the other.
Answer. $89760=9 \cdot 10^{4}-2 \cdot 5!$.

Solution. We will count the number of five-digit numbers in which no digit is divisible by any other digit, and then subtract this number from the total number of five-digit numbers, which is 90000, to get the answer to the problem.

Note that a number, none of whose digits are divisible by any other, does not contain the digit 0 (0 is divisible by any digit), 1 (any digit is divisible by 1), and does not contain identical digits (they always divide each other). Therefore, the smallest digit of such a five-digit number can be $2, 3, 4$ or 5. If it is 2, then the other four digits must be odd, specifically $3, 5, 7$ and 9, but 9 is divisible by 3. If it is 3, the other four digits can only be $4, 5, 7, 8$, but 8 is divisible by 4. The only numbers left are those composed of the digits $4, 5, 6, 7, 9$ or $5, 6, 7, 8, 9$. In each case, there are $5!=120$ possible permutations of these digits, totaling 240 numbers. Therefore, the number of numbers sought in the problem is 90000 $- 240 = 89760$.

Grading criteria. The idea of applying the principle of subtraction: 3 points. Correct calculation of the number of five-digit numbers: 1 point. Correct calculation of the number of five-digit numbers in which no digit is divisible by any other digit: 3 points. If the solution does not account for the divisibility of 0 by any digit or the divisibility of any digit by 1 or equal digits by each other: minus 3 points.