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Math Help - Arrangements with letters

  1. #1
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    Arrangements with letters

    im having alot of trouble with these letter questions

    How many arrangements can be made by the letters of the word definition if the letters i do not occupy the first or last place
    and how many arrangements of the letters of the word PARRAMATTA are possible

    i dont no how to approach that first question and for the second one i thought it was just 10!/8 but the answer is 37800

    thankyou for any help given.
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  2. #2
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    PARRAMATTA There are 10 ltters total. There are 4 A's, 2 T's, and 2 R's.

    \frac{10!}{4!2!2!}
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  3. #3
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    Hello, flyinhigh123!

    The first one is tricky.
    I took me while to come up with an approach.


    How many arrangements can be made by the letters of the word DEFINITION
    if the letter I does not occupy the first or last place?
    There are 10 letters: . D, E, F, III, NN, O, T . . . with three Is and two Ns.


    There are 10 positions: . _ _ _ _ _ _ _ _ _ _


    There are 8 "interior" positions.
    . . Choose 3 of them for the Is.
    . . There are: . {8\choose3} choices.

    The remaining 7 letters \{D,E,F,N,N,O,T\} can be placed
    . . in the remaining 7 positions in: . \frac{7!}{2!} ways.


    Therefore, there are: . {8\choose3}\cdot\frac{7!}{2!} \:=\:56\cdot 2520 \:=\:141,\!120 arrangements.

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  4. #4
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    Quote Originally Posted by flyinhigh123 View Post
    im having alot of trouble with these letter questions

    How many arrangements can be made by the letters of the word definition if the letters i do not occupy the first or last place
    and how many arrangements of the letters of the word PARRAMATTA are possible

    i dont no how to approach that first question and for the second one i thought it was just 10!/8 but the answer is 37800

    thankyou for any help given.
    You need to use the "fundamental counting principle": "If A can take place in n ways and B can take place in m ways then A and B can take place in mn ways."

    "How many arrangements can be made by the letters of the word 'definition' if the letters i do not occupy the first or last place."
    (The quotes around 'definition' are important here. Otherwise, I'm wondering what 'definition' you are talking about!)
    Since the first letter cannot be an "i", there are 8 letters that can be be first- there are 8 choices for the first letter. Since the last letter cannot be an "i", and we have already used one of the 8 other letters, there are now 7 choices for the last letter. For each of the 8 positions, we can now use the "i"s so there are now 8 choices for the second letter, 7 for the third etc. That gives a total of (8)(7)(8)(7)(6)(5)(4)(3)(2)(1).

    But hold on, we aren't done yet! That is treating the 3 "i"s and the 2 "n"s as if they were "distinguishable". They aren't: swapping the theee "i"s and/or the 2 "n"s while leaving the other other letters alone would give exactly the same word. Since there are 3!= 6 ways to swap the "i"s and 2!= 2 ways to swap the "n"s, that previous product is 12 times as large as it should be. We fix that by dividing by 4, of course. The number of (distinguishable) ways of rearranging the letters in "definition" is (8)(7)(8)(7)(6)(5)(4)(3)(2)(1)/12= (8)(7)(8)(7)(6)(5)(4)(3)(2)(1)/(4)(3)= (2)(7)(8)(7)(6)(5)(4)(3)(2)(1) which can also be written as (8!)(14/3).

    You can do much the same thing with "PARRAMATTA". There are a total of 10 letters so if they were all different there would be 10! ways to interchange them. But 4 of those letters are indistinguishable "A"s , 2 are "R", and 2 are "T" so the actual number of different permutations is 10!/(4!)(2!)(2!)= 37800. You treated all 8 "A, R, T"s as if they were the same.
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  5. #5
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    thankyou galactus, soroban and hallsofivy for ALL your help ^^ i really, i realli appreciate it, and your explanations are awesum, you guyz are completely awesome =D
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