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A 5 digit number is selected at random. The probability that the chosen number has distinct digits and digits in the odd places are odd and digits in the even places are even is

Options

  1. A5 P 2 × 5 P 3 9 × 1 0 4
  2. B5 P 2 × 5 P 3 1 0 5
  3. C5 C 2 × 5 C 3 × 2 1 0 4 × 9
  4. D5 C 2 × 5 C 3 9 × 1 0 4

Correct answer

A. 5 P 2 × 5 P 3 9 × 1 0 4

Step-by-step solution

Total cases = 9 × 1 0 4 Favourable cases = 50 = 5 P 3 × 5 P 2 Hence, required probability = 5 P 3 × 5 P 2 9 × 1 0 4

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