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A five digit number is choosen 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?
Asked by laks78 | 13 May, 2019, 23:39: PM
answered-by-expert Expert Answer
Total numbers = 9 × 104 
Digits at odd places are odd and at even places is even.
3 odd places 1, 3, 5, 7, 9 and 2 even places contain 0, 2, 4, 6, 8
Total number of ways = 5 × 4 × 3 × 5 × 4 = 1200
Required probability = 1200/90000 = 1/75
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