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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)
    9/56
  • b)
    4/9
  • c)
    3/7
  • d)
    11/27
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4,...
Total cases:

Favourble 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!
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Community Answer
Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4,...
Total cases:

Favourble 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!
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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)9/56b)4/9c)3/7d)11/27Correct answer is option 'B'. Can you explain this answer?
Question Description
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