Problem:
The number of 3-digit even numbers that can be formed from the digits 0, 1, 2, 3, 4, and 5, repetition of digits being not allowed.

Solution:
To find the number of 3-digit even numbers that can be formed, we need to consider the following factors:

1. The first digit cannot be zero since it would result in a two-digit number.
2. The last digit must be an even number (0, 2, 4) since we are looking for even numbers.
3. The second digit can be any of the remaining four digits (1, 2, 3, 5) since repetition is not allowed.

Step 1:
Counting the possibilities for the first digit:
- We have five options (1, 2, 3, 4, 5) for the first digit since zero is not allowed.
- Therefore, there are 5 possibilities for the first digit.

Step 2:
Counting the possibilities for the second digit:
- Since repetition is not allowed, the second digit can be any of the remaining four digits (1, 2, 3, 5).
- Therefore, there are 4 possibilities for the second digit.

Step 3:
Counting the possibilities for the third digit:
- The third digit must be an even number (0, 2, 4).
- Therefore, there are 3 possibilities for the third digit.

Step 4:
Combining the possibilities from Steps 1, 2, and 3:
- To find the total number of possibilities, we multiply the number of possibilities for each step.
- 5 possibilities for the first digit x 4 possibilities for the second digit x 3 possibilities for the third digit = 60 possibilities.

Final Answer:
Therefore, the correct answer is option 'C' as there are 60 different 3-digit even numbers that can be formed from the given digits without repetition.