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by Atlas on December 8th, 2009

Atlas

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Let S3 be the six digit numbers whose one's digit equals the number of 3's appearing in the number. find the number of S(3) numbers

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  • by Quirkie on December 9th, 2009

    Quirkie

    It doesn't seem worth making a special formula. Just look at each in turn:
    Last digit = 0, 8*9*9*9*9*1 numbers.
    Last digit = 1, first digit 3: 1*9*9*9*9*1
    Last digit = 1, first digit not 3: 8*1*9*9*9*1 * 4
    Last digit = 2, first digit 3: 1*1*9*9*9*1 * 4
    Last digit = 2, first digit not 3: 8*1*1*9*9*1 * 6
    Last digit = 3, first digit 3: 1*1*9*9*9*1 * 4
    Last digit = 3, first digit not 3: 8*1*1*9*9*1 * 6
    Last digit = 4, first digit 3: 1*1*1*1*9*1 * 4
    Last digit = 4, first digit not 3: 8*1*1*1*1*1
    Last digit = 5, first digit 3: 1*1*1*1*1*1

    And just add them up.

    In the above, the first six factors in, for example, 1*1*1*1*1*1 are the possibilities for each digit, and the extra factor is the number of ways of arranging the 3s given the conditions.

    This would be significantly simpler if the first digit of a six digit number is permitted to be 0

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