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Math Help - Counting Principles, five-digit number combinations

  1. #1
    DBA
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    Counting Principles, five-digit number combinations

    Hello,

    How many even five-digit numbers are there?

    The answer is 5*9*10^3

    I have 5 places _ _ _ _ _
    I thought for a number to be even, the last place needs to contain an even number (0,2,4,6,8) which are 5 possibilities.

    The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.

    The following part of the question is

    How many five-digit numbers are there with exactly one 3?
    Here I do not even know where to start.

    Thanks for any help
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    Quote Originally Posted by DBA View Post

    The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.
    If the first digit is 0, we do not consider it a 5 digit number any longer. (Usually stated as, disallow leading zeroes.)

    Quote Originally Posted by DBA View Post
    The following part of the question is

    How many five-digit numbers are there with exactly one 3?
    Here I do not even know where to start.

    Thanks for any help
    You can view it as one of 3xxxx, x3xxx, xx3xx, xxx3x, xxxx3 where x's are blanks (easier to type x).
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  3. #3
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    Quote Originally Posted by DBA View Post
    How many even five-digit numbers are there?
    The answer is 5*9*10^3
    I have 5 places _ _ _ _ _

    The answer indicates that some other place only can have 9 choices, instead of 10 (0-9). I do not understand why.

    How many five-digit numbers are there with exactly one 3?
    Here I do not even know where to start.
    The reason for the 9 is your text does not allow leading zeros.
    i.e 04542 does not count as a five-digit number.

    For your second one 9^4 is number with a 3XXXX in the lead.
    Then 8\cdot 4\cdot 9^3, eight choices for the lead digit, four choices for the 3 and choosing the three other digits.

    If we add those together, 9^3\cdot 41.
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  4. #4
    DBA
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    I looked at every position:

    1st position: All positions cannot contain a three
    3XXXX 9*9*9*9=9^4

    2nd position: First position cannot be zero and all positions cannot contain a three
    X3XXX 8*9*9*9=8*9^3

    3rd position:
    XX3XX 8*9*9*9=8*9^3

    4th position:
    XXX3X 8*9*9*9=8*9^3

    5th position:
    XXXX3 8*9*9*9=8*9^3

    They are disjoint events to me, so I would add them up:

    9^4 + 8*9^3 + 8*9^3 + 8*9^3 + 8*9^3 = 9^4 + 4(8*9^3)

    Thanks! The answer in the book says 9^4 + (4*8*9^3)
    Last edited by DBA; September 24th 2010 at 11:24 AM. Reason: additional information about answer
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