Official solution
Total cases :
\[\underline 6 \] . \[\underline 6 \] . \[\underline 5 \] . \[\underline 4 \] . \[\underline 3 \] . \[\underline 2 \]
n(s) = 6 . 6!
Favourable cases :
Number divisible by 3 \[ \equiv \] Sum of digits must be divisible by 3
Case - I
1, 2, 3, 4, 5, 6
Number of ways = 6!
Case - II
0, 1, 2, 4, 5, 6
Number of ways = 5 . 5!
Case - III
0, 1, 2, 3, 4, 5
Number of ways = 5 . 5!
n(favourable) = 6! + 2 . 5 . 5!
\[P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}\]