A five digit number is chosen at random. The probability that all the digits are distinct and digits at odd places are odd and digits at even places are even is
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Answer :- 1/75
Total numbers = 9 × 10 × 10 × 10 × 10
digit at odd place are odd and at even places are even.
∴ 3 odd places contains 1, 3, 5, 7, 9, ways and 2 even places contain 0, 2, 4, 6, 8.
∴ Total numbers of ways = [5 × 4 × 3] × [5 × 4] = 1200
∴ probability = [(1200)/(9 × 10000)] = (1/75)
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