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Math Help - Four-Digit Odd Numbers

  1. #1
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    Four-Digit Odd Numbers

    How can I solve this problem without using the fundamental counting principle?

    How many 4-digit odd numbers greater than 3,000 can be formed by using the digits 0-9 if digits may be repeated?

    I was told this is a permutation NOT a combination word problem.

    In that case, can I apply nPn or nPr. If not, why not?
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  2. #2
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    Hello, sharkman!

    How many 4-digit odd numbers greater than 3,000 can be formed
    by using the digits 0-9 if digits may be repeated?
    We have a four-digit number.

    The first (leftmost) digit can be any of {3, 4, 5, 6, 7, 8, 9} . . . 7 choices.

    The second digit can be any of the 10 digits.

    The third digit can be any of the 10 digits.

    The last (rightmost) digit can be any of {1, 3, 5, 7, 9} . . . 5 choices.


    Therefore . . .

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  3. #3
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    Then...

    Quote Originally Posted by Soroban View Post
    Hello, sharkman!

    We have a four-digit number.

    The first (leftmost) digit can be any of {3, 4, 5, 6, 7, 8, 9} . . . 7 choices.

    The second digit can be any of the 10 digits.

    The third digit can be any of the 10 digits.

    The last (rightmost) digit can be any of {1, 3, 5, 7, 9} . . . 5 choices.


    Therefore . . .
    7*10^2*5 = 3500


    Is there a way to solve this using a formula?
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