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Math Help - [SOLVED] Permutations Problem

  1. #1
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    [SOLVED] Permutations Problem

    The language with the most letters in it is Cambodian, which has 72 letters.

    b) How many three-letter permutations can be created if the letters may be repeated?



    The options would look something like this: AAABBBCCCDDDEEEFFF...

    So I tried: \frac{72!}{72(3!)} but that was too big for my calculator.

    I don't know how else to solve this.


    Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by Some_One View Post
    The language with the most letters in it is Cambodian, which has 72 letters.

    b) How many three-letter permutations can be created if the letters may be repeated?



    The options would look something like this: AAABBBCCCDDDEEEFFF...

    So I tried: \frac{72!}{72(3!)} but that was too big for my calculator.

    I don't know how else to solve this.


    Any help is greatly appreciated.
    ^nP_r = \frac{n!}{(n-r)!}

    n = 72 and r = 69

    ^nP_r = \frac{72!}{(72-3)!} = (72) \times (71) \times (70)
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  3. #3
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    Hello, Some_One!

    The language with the most letters in it is Cambodian, which has 72 letters.

    b) How many three-letter permutations can be created if the letters may be repeated?


    The options would look something like this: AAABBBCCCDDDEEEFFF... . what?

    So I tried: \frac{72!}{72(3!)} .??
    These are three-letter "words" like: . AAA, AAB, AAC\; \hdots\; ZZZ

    There are 72 choices for the first letter.
    . . There are 72 choices for the second letter.
    . . . . There are 72 choices for the third letter.

    Therefore . . .

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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    ^nP_r = \frac{n!}{(n-r)!}

    n = 72 and r = 69

    ^nP_r = \frac{72!}{(72-3)!} = (72) \times (71) \times (70)
    That would be if the letters may NOT be repeated.

    My problem is if letter CAN be repeated.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Some_One!

    These are three-letter "words" like: . AAA, AAB, AAC\; \hdots\; ZZZ

    There are 72 choices for the first letter.
    . . There are 72 choices for the second letter.
    . . . . There are 72 choices for the third letter.

    Therefore . . .


    Doh! Of course! I can't believe that I forgot that.


    Before this question I was doing a problem that involved finding the permutations of A,B,D,D,D so I still had that mind set.


    Thanks for your help!
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  6. #6
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    Hello again, Some_One!

    Before this question I was doing a problem that involved finding
    the permutations of A,B,D,D,D so I still had that mind set.
    Boy, do I know that feeling!

    I have this flat spot on my forehead to prove it.

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