How many three-digit numbers are there in which the tens digit is greater than the ones digit?

Answer 1

Considering only positive integers then:

We are looking at all the numbers between 100 to 999 (inclusive).

Or 9 sets of 100.

For each set of 100 (e.g. 100 to 199 or 200 to 299)

There will be 0+1+2+3+4+5+6+7+8+9 = 45 numbers where this statement holds true.

We know this because for 100 to 109 there are none.

For 110 to 119 there is 1 (110)

For 120 to 129 there are 2 (120 & 121)

and this pattern holds all the way to 190 to 199 where there are 9.

This will be true for each set of 100 numbers so the answer is 45 * 9 = 405.