Question

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 :

• A. $\frac{9}{56}$
• B. $\frac{4}{9}$
• C. $\frac{3}{7}$
• D. $\frac{11}{27}$

Answer 1

Answer: (B)

Total cases:
$\underline{6} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2}$
$n ( s )=6 \cdot 6 !$
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 $=5 \cdot 5 !$
Case-III
$0,1,2,3,4,5$
Number of ways $=5.5 !$
$n ($ favourable $)=6 !+2 \cdot 5 \cdot 5 !$
$P =\frac{6 !+2 \cdot 5 \cdot 5 !}{6 \cdot 6 !}=\frac{4}{9}$