How many numbers arc there between 99 and 1000 such that the digit 8 occupies the units place?

The numbers between 99 and 1000 are three-digit numbers, ranging from 100 to 999. We need to find numbers in this range where the units digit is 8.

Let the number be represented as H T U, where H is the hundreds digit, T is the tens digit, and U is the units digit.

1. Units digit (U): It must be 8. So, there is only 1 choice for the units digit.
2. Hundreds digit (H): Since the numbers are between 99 and 1000, they are three-digit numbers (100 to 999). The hundreds digit cannot be 0. So, H can be any digit from 1 to 9 (1, 2, 3, 4, 5, 6, 7, 8, 9). This gives 9 choices.
3. Tens digit (T): The tens digit can be any digit from 0 to 9 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). This gives 10 choices.

To find the total number of such numbers, we multiply the number of choices for each digit: Total numbers = (Choices for H) * (Choices for T) * (Choices for U) Total numbers = 9 * 10 * 1 = 90.

Therefore, there are 90 numbers between 99 and 1000 such that the digit 8 occupies the units place.

The final answer is C.