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Math Help - Counting Identical Elements

  1. #1
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    Counting Identical Elements

    How many arrangements of letters of word TRANSITION are possible if
    the Is and Ns and Ts are together.

    Here is my attempt:
    Group Is, Ns, and Ts together, and arrange: 5! ways
    Arrange amonst Is, Ns, Ts, removing repeition: 6! / 2!2!2!

    This is not what the book has, which contains 5040 as answer.
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    How many arrangements of letters of word TRANSITION are possible if
    the Is and Ns and Ts are together.

    Here is my attempt:
    Group Is, Ns, and Ts together, and arrange: 5! ways
    Arrange amonst Is, Ns, Ts, removing repeition: 6! / 2!2!2!

    This is not what the book has, which contains 5040 as answer.
    When you require that for example the T's be together, that means you treat it as a single entity, so since there are 7 distinct letters, the answer is simply 7! as the book says.
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  3. #3
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    Hello, Lukybear!

    undefined is absolutely correct.


    \text{How many arrangements of letters of word TRANSITION}
    \text{ are possible if  the I's and N's and T's are together?}

    Since the double-letters are to be adjacent, duct-tape them together.

    Then we have 7 "letters" to arrange: . \bigg \{A,\;O,\;R,\;S,\;\boxed{II},\;\boxed{NN},\;\boxed  {TT}\bigg \}

    Therefore, there are: . 7! \,=\,5040 arrangements.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I see another interpretation of this problem.

    It should have said: . \begin{Bmatrix}\text{"The I's are adjacent,} \\ \text{and the N's are adjacent,} \\ \text{and the T's are adjacent."} \end{Bmatrix}


    Instead it said: "The I's and N's and T's are together."
    . . which could mean that we have, for example, NTINIT
    . . somewhere in the arrangement.

    So we have 5 "letters" to arrange: . \bigg\{A,\:O,\:R,\:S,\:\boxed{IINNTT} \bigg\}
    . . There are: . 5!= 120 ways.

    But \boxed{IINNTT} can be arranged in: . \displaystyle {6\choose2,2,2} = 90 \text{ ways.}


    Therefore, there are: . 120 \times 90 \:=\:10,800 arrangements.

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