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Math Help - Forming 3-digit numbers

  1. #1
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    Forming 3-digit numbers

    The digits 3, 6, 7, and 8 will be used without repetition to form different 3-digit numbers. Of all such numbers, how many are greater than 500?

    The answer is 18 but don't know how to obtain it
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  2. #2
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    Re: Forming 3-digit numbers

    3-digit numbers greater than 500 must not start with 3 so = 4P3 - 3P2 = 18
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  3. #3
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    Re: Forming 3-digit numbers

    I'm still confused--I understand why they cant start with 3 (needs to be more than 500) but I don't understand the repetition part. Also, what does the last part mean with the "P"
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  4. #4
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    Re: Forming 3-digit numbers

    Hello, dannyc!

    The digits 3, 6, 7, and 8 will be used without repetition
    to form different 3-digit numbers.
    Of all such numbers, how many are greater than 500?

    The answer is 18 but don't know how to obtain it.

    I don't know if we can help you.

    This is definitely a Permutation problem.
    . . But you've never heard of Permutations, _nP_r ?

    What grade are you in?

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  5. #5
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    Re: Forming 3-digit numbers

    HINT:
    367,368,376,378,386,387 ; there's 3 more similar arrangements...

    Btw, no repetition means no number like 777 or 778; all digits must be different...

    Go here to "learn":
    Math Forum: Ask Dr. Math FAQ: Permutations and Combinations
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  6. #6
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    Re: Forming 3-digit numbers

    In high school Soroban and yes have done plenty of permutations--now I know what the "P" is from the post above but not sure why to choose those numbers to form the permutation.
    Wimer--the concrete numbers like that help, would the 3 other arrangements consist of numbers starting with 6, 7, and 8? (that is, numbers in the 600s, 700s, and 800s?)
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  7. #7
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    Re: Forming 3-digit numbers

    Quote Originally Posted by dannyc View Post
    Wimer--the concrete numbers like that help, would the 3 other arrangements consist of numbers starting with 6, 7, and 8? (that is, numbers in the 600s, 700s, and 800s?)
    Of course!
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  8. #8
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    Re: Forming 3-digit numbers

    So then there would be 6 numbers to those 3 sets as well? (637,638,673,678,683,687) Would those be the correct ones for the 600s?
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  9. #9
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    Re: Forming 3-digit numbers

    The first number can be either 6, 7, or 8- three choices. The second number can be either of the two remaining digits and 3- three choices again. There are then 2 choices for the third digit. A total of 3*3*2= 18 possible numbers.
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  10. #10
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    Re: Forming 3-digit numbers

    Thanks Ivy!! That makes much more sense! (I should've thought of that)
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  11. #11
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    Re: Forming 3-digit numbers

    Quote Originally Posted by dannyc View Post
    So then there would be 6 numbers to those 3 sets as well? (637,638,673,678,683,687) Would those be the correct ones for the 600s?
    Danny boy, why ain't you SURE those are correct?
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  12. #12
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    Re: Forming 3-digit numbers

    LOL I guess I just needed that confidence boost. Thanks!
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