Permutations Practice Test 1

We know that a number is divisible by 3 if the sum of its digits is divisible by 3.
Now the sum of the digits 1, 2, 3, 4 and 5 is 15 which is divisible by 3.
∴ All the five digits number formed by the digits 1, 2, 3, 4, 5 are divisible by 3 and their number = 5! = 120
When we include 0, the four other digits whose sum is divisible by 3 are 1, 2, 4 and 5
The number of numbers in this case = 4 × 4! = 4 × 24 = 96
Hence, the required number of numbers = 120 + 96 = 216