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 :

Step-by-step solution

Total cases : 6 · 6 · 5 · 4 · 3 · 2 n s = 6 · 6   ! Favourable cases : Number divisible by 3 ≡ 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   ! 6 · 6   ! = 4 9