How many different 3-digit even numbers can be formed so that all the digit are different?

To solve this problem, we need to consider the three places in the 3-digit number separately: the hundreds place, the tens place, and the units place.

1. Units Place: Since the number has to be even, the units place can be filled by any one of the 5 even numbers from 0 to 9, which are 0, 2, 4, 6, and 8. So, there are 5 ways to fill the units place.

2. Hundreds Place: The hundreds place can't be 0 (because then it wouldn't be a 3-digit number) and it can't be the same as the units place. So, there are 8 remaining choices (1-9, excluding the number chosen for the units place). So, there are 8 ways to fill the hundreds place.

3. Tens Place: The tens place can be any of the remaining 8 digits (0-9, excluding the numbers chosen for the hundreds and units places). So, there are 8 ways to fill the tens place.

So, the total number of different 3-digit even numbers that can be formed so that all the digits are different is 5 (units place) * 8 (hundreds place) * 8 (tens place) = 320.