Verified Solution • 2025–26 Session moderate

QJEEM21Maa09b: JEE Main 2021 Mathematics - Class 12 JEEM21Maa09b Solution: Let A denote the event that a 6-digit integer formed by 0, 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 :

Multiple Choice Options

  • A 49{4 \over {9}}
  • B 956{9 \over {56}}
  • C 1127{11 \over {27}}
  • D 37{3 \over {7}}

Step-by-Step Verified Solution

Total cases :

6‾\underline 6 . 6‾\underline 6 . 5‾\underline 5 . 4‾\underline 4 . 3‾\underline 3 . 2‾\underline 2

n(s) = 6 . 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 . 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!=49P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}

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