While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred's place is greater than the digit at ten's place, and the digit at ten's place is greater than the digit at unit's place?

- 61
- 64
- 85
- 91

### Answer

For numbers from 700 to 1000, the digit at hundred's place can be 7, 8 or 9. So, three possibilities are there.

Now, for the digit at ten's place to be greater than the digit at unit's place, total numbers is equal to the value of ten's digit. For example, for 2 at ten's digit, numbers can be 20, 21 (total 2). Similarly, for 6 at ten's place, numbers can be 60, 61, 62, 63, 64, 65 (total 6).

When the first digit is 7, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 = 21

When the first digit is 8, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

When the first digit is 9, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Therefore, required total numbers = 21 + 28 + 36 = 85.

**The correct option is C.**