Total digit = 6 {0, 1, 2, 3, 4 and 5}
We have to formed 3-digit even numbers:
Case-1: When 0 is at unit place, then
Number of choices for the units place = 1
(0 is at unit place)
Number of choices for the ten's place = 5
(Any digit except the one that is already used as the 1st digit)
Number of choices for the hundredth's place = 4
(Any digit except the ones that are already used as the 1st and 2nd digit)
No. of such three digit even number = 1 × 5 × 4 = 20
Case-2: When 0 is at ten's place, then
For even number unit place must be filled with 2 or 4
Number of choices for the units place = 2
Number of choices for the ten's place = 1
Number of choices for the hundredth place = 4
No. of such three digit even number = 2 × 1 × 4 = 8
Case-3: When no digit in the 3-digit even number is zero.
For even number unit place must be filled with 2 or 4
Number of choices for the units place = 2
Number of choices for the ten's place = 4
Number of choices for the hundredth place = 3
No. of such three digit even number = 2 × 4 × 3 = 24
Total number of possible three digit even number = 8 + 24 + 20 = 52