1. How many 4-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place?
Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: 1, $2,3, \ldots, 9$. That is, there are a total of 45 options. The other two digits are arbitrary (10 options each), so we get the answer: $45 \cdot 10 \cdot 10=4500$ options.