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Math Help - combinatorics problem involving even numbers?

  1. #1
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    combinatorics problem involving even numbers?

    The problem is: How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6 and 9 if each digit must be used exactly once in each number?

    Like all combinatorics problems, I create five spots. The last spot must be used up by 4 or 6, b/c it's even, so we have

    _ _ _ _ 2

    But then my book says to assign the first spot to 3, because there are 3 numbers left over after choosing two for the last spot. But since I only used one number for the last spot--either 4 or 6---shouldn't the first spot be assigned to 4, and then, accordingly:

    4 3 2 1 3 ?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by lu6cifer View Post
    The problem is: How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6 and 9 if each digit must be used exactly once in each number?

    Like all combinatorics problems, I create five spots. The last spot must be used up by 4 or 6, b/c it's even, so we have

    _ _ _ _ 2

    But then my book says to assign the first spot to 3, because there are 3 numbers left over after choosing two for the last spot. But since I only used one number for the last spot--either 4 or 6---shouldn't the first spot be assigned to 4, and then, accordingly:

    4 3 2 1 3 ?
    The key here is "greater than 40'000" - the first number CANNOT be that 3. Thus we have 3 choices for the first number. So we have,

    3 3 2 1 2

    which gives us 36 choices.
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  3. #3
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    ooo..you're right

    must've forgotten that part amidst all the combinatorics stuff.
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