Official solution
Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.
Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.
Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)
So total no of arrangement = 1\[ \times \]2\[ \times \]3\[ \times \]4\[ \times \]5 = 5!
Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)
So total no of arrangement = 4\[ \times \]4\[ \times \]3\[ \times \]2\[ \times \]1 = 4.4!
\[\therefore\] Total arrangement = 5! + 4.4! = 216