JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
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 :
Probability
Solution
Six distinct digits from $0..6$ omit one digit; sum of all is 21, so divisible by 3 iff omitted digit is $0,3$ or $6$. Omit 0: $6!=720$; omit 3 or 6: $6!-5!=600$ each. Favourable $=1920$. Total 6-digit numbers: $6\cdot6!=4320$ ($720+6\cdot600$). $P=\frac{1920}{4320}=\frac49$. Option (A).