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

The correct answer is Option 90.

We are asked to find the total number of integers between 99 and 1000 such that the digit 8 occupies the units (ones) place.

Step 1: Determine the range of numbers
The numbers strictly between 99 and 1000 range from 100 to 999 inclusive.
Notice that all numbers in this range are 3-digit numbers.

Step 2: Understand the place value requirements
Let a 3-digit number be represented as:
$HTU$
where:
- $H$ is the Hundreds digit.
- $T$ is the Tens digit.
- $U$ is the Units digit.

Step 3: Analyze the possibilities for each digit position
1. Units place ($U$):
According to the problem, the units place must be occupied by the digit 8.
Therefore, there is only 1 choice for the units place ($U=8$).

2. Hundreds place ($H$):
The hundreds digit can be any digit from 1 to 9 (since a 3-digit number cannot start with 0).
Digits available: {1, 2, 3, 4, 5, 6, 7, 8, 9}.
So, there are 9 possible choices for the hundreds place.

3. Tens place ($T$):
The tens digit can be any digit from 0 to 9.
Digits available: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
So, there are 10 possible choices for the tens place.

Step 4: Calculate the total number of such numbers
Using the Fundamental Counting Principle, the total number of 3-digit numbers with 8 in the units place is:

$\text{Total numbers}=\left(\text{Choices for}H\right)\times \left(\text{Choices for}T\right)\times \left(\text{Choices for}U\right)$

$\text{Total numbers}=9\times 10\times 1=90$

Thus, there are 90 such numbers between 99 and 1000 having the digit 8 in their units place.