Between 10 and 99, how many numbers are there such that, by reversing the order of their digits, we obtain a number greater than the original number?
We should count the two-digit numbers $ab$ where the unit digit $b$ is greater than the tens digit $a$, that is, such that $b>a$. If $a=1$, the unit digit $b$ can be $2,3,4,5,6,7,8$ or 9, so we have eight possibilities. If $a=2$, the unit digit $b$ can be $3,4,5,6,7,8$ or 9, so we have seven possibilities. Continuing in this way, we reach $a=8$, when the unit digit $b$ can only be 9, so we have only one possibility. Clearly, $a$ cannot be 0 or 9. Therefore, there are $8+7+6+5+4+3+2+1=36$ numbers between 10 and 99 such that, by reversing the order of their digits, we obtain a number greater than the original number.