How many two-digit numbers are there where the digit in the tens place is greater than the digit in the units place?

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The problem can be solved by simple enumeration.

1) If the units digit is 0, then the tens digit can take values from 1 to 9.
2) If the units digit is 1, then the tens digit can take values from 2 to 9.
3) If the units digit is 2, then the tens digit can take values from 3 to 9.
4) If the units digit is 8, then the tens digit can take only the value 9.

Therefore, the total number of cases is $1+2+3+\ldots+9=45$ numbers.