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 number of 6 digited numbers from0, 1, 2, 3, 4, 5, 6          =  6×6×5×4×3×2       The number of events in the sample space is nS=6×6!Therefore 6 digited number divisible by 3 can be formed by using the digits           1,2,3,4,5,6 in 6!ways  by using the digits 0,1,2,4,5,6 in 5×5! ways and           by using the digits 0,1,2,3,4,5 in  5×5! ways PE=6!+2×5×5!6×6!=6+1036=49