There are 9999 tickets bearing numbers 0001, 0002, … , 9999. if one ticket is selected from these tickets at random, the probability that thenumber on the ticket will consists of all different digits, is

There are 9999 tickets bearing numbers 0001, 0002, ... , 9999. if one ticket is selected from these tickets at random, the probability that the
number on the ticket will consists of all different digits, is

  1. A

    50409999

  2. B

    50009999

  3. C

    50309999

  4. D

    None of these

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    Solution:

    Total number of cases = 9999
    Favorable cases = 10 x 9 x8 x 7 = 5040
    Probability =50409999

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