Correct Option (3)
The problem requires identifying the count of numbers between 99 and 1000 that have the digit 8 in the units place. Numbers between 99 and 1000 are three-digit numbers, ranging from 100 to 999.
For a three-digit number represented as H T U (Hundreds, Tens, Units):
- The Units place (U) must be 8. This provides 1 possibility.
- The Hundreds place (H) cannot be 0, as it would then be a two-digit number. Therefore, H can be any digit from 1 to 9. This provides 9 possibilities.
- The Tens place (T) can be any digit from 0 to 9. This provides 10 possibilities.
Using the fundamental principle of counting, the total number of such numbers is the product of the possibilities for each position:
Total numbers = (Possibilities for H) × (Possibilities for T) × (Possibilities for U)
Total numbers = 9 × 10 × 1 = 90.
Thus, there are 90 numbers between 99 and 1000 where the digit 8 occupies the units place.
Incorrect Options:
Options 1 (64), 2 (80), and 4 (104) are incorrect. These values do not result from the correct application of combinatorial principles for determining the number of three-digit integers with the specified constraints. The calculation based on the fixed units digit and the permissible range for the hundreds and tens digits yields 90 as the accurate count.