Permutations Practice Test 1

The number is divisible by 3 if the sum of all its digits is divisible by 3.
Here sum of all six digits is 15. The sum of five digits will also be divisible by 3 if we leave either 0 or 3 from the six digits.
Leaving 3, total no. of numbers = 5! – 4!.
(Since the number having 0 as first place are of four digits like 01245).
And leaving 0, total no. of number = 5!.
Hence, the required number = (5! – 4!) + 5! = 216.