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The digits 2, 4, 6, 8 and 0 are used to make five-digit numbers with no digits repeated. What is the probability that a number chosen at random from these numbers has the property that the digits in the thousands place and ten's place are each larger than their neighboring digits.
a)5/48
b)11/96
c)1/8
d)13/96
e)7/48
So I know that it's out of 4 * 4! combinations because you can't have the zero as the first digit, how do I find how many numbers meet the condition?

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Alex

  1. Case1: the 6 in in thousand place and 8 in the tens place, leaving the 0,2,4
    _6_8_ -----> 2x2x1 = 4
    Case2:
    _8_6_ -----> 2x2x1 = 4
    Case3:
    _8_4_ , the 6 must be the lead digit
    68_4_ ------> 2x1 = 2

    a total of 10
    prob(your event) = 10/96 = 5/48

    it would be simple to actually list the numbers

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    Reiny

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