$$Let $A$ denote the event that a $6$-digit integer formed by $0,1,2,3,4,5,6$ without repetitions, be divisible by $3$ . Then probability of event $A$ is equal to :

Total cases : $\underline{6}·\underline{6}·\underline{5}·\underline{4}·\underline{3}·\underline{2}$

$n\left(s\right)=6·6 !$

Favourable cases :

Number divisible by $3\equiv$ Sum of digits must be divisible by $3$

Case-$\mathrm{I}$

$1,2,3,4,5,6$

Number of ways $=6 !$

Case-$\mathrm{II}$

$0,1,2,4,5,6$

Number of ways $=5·5 !$

Case-$\mathrm{III}$

$0,1,2,3,4,5$

Number of ways $=5·5 !$

$\mathrm{n}($ favourable $)=6 !+2·5·5 !$

$P=\frac{6 !+2·5·5 !}{6·6 !}=\frac{4}{9}$