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Math Help - Another easy combinatorics problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Another easy combinatorics problem

    How many three-digit numbers are there in which the sum of the digits is even?

    There are 900 three-digit numbers.
    I guess half of them (450) will have even sum of digits. Is this right? I don't know how to prove this, though.
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  2. #2
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    Re: Another easy combinatorics problem

    Quote Originally Posted by alexmahone View Post
    How many three-digit numbers are there in which the sum of the digits is even?
    There are 900 three-digit numbers.
    If three integers add to an even then all three are even or one even and two are odd. Count them.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Another easy combinatorics problem

    Quote Originally Posted by Plato View Post
    If three integers add to an even then all three are even or one even and two are odd. Count them.
    All three digits are even: 4*5*5 = 100
    First digit is even, second and third are odd: 4*5*5 = 100
    First digit is odd, second is even, third is odd: 5*5*5 = 125
    First digit is odd, second is odd, third is even: 5*5*5 = 125

    Total: 100 + 100 + 125 + 125 = 450
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    Re: Another easy combinatorics problem

    Quote Originally Posted by alexmahone View Post
    All three digits are even: 4*5*5 = 100
    First digit is even, second and third are odd: 4*5*5 = 100
    First digit is odd, second is even, third is odd: 5*5*5 = 125
    First digit is odd, second is odd, third is even: 5*5*5 = 125

    Total: 100 + 100 + 125 + 125 = 450
    Correct, half of them give an even sum.
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