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Math Help - Tricky Permutation/Graph theory problem

  1. #1
    Member elninio's Avatar
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    Tricky Permutation/Graph theory problem

    Count the permutations θ in S_6 that satisfy θ(1)=2 and θ(2)=3.

    I'm mostly confused by the wording of the problem. Would it be implying that a permutation can ONLY contain those criteria OR that it can contain those two and any other subsequent "run".
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  2. #2
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    Quote Originally Posted by elninio View Post
    Count the permutations θ in S_6 that satisfy θ(1)=2 and θ(2)=3.
    If I were given this problem as written, I would answer 4!.
    That is the number of ways to assign 3,4,5,~\&~6.
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  3. #3
    Member elninio's Avatar
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    My understanding is that the permutations do no have to include 3,4,5 and 6.
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  4. #4
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    Quote Originally Posted by elninio View Post
    My understanding is that the permutations do no have to include 3,4,5 and 6.
    This may well be a case where notation not being standard is the difficulty.
    The symbol S_6 usually stands for the symmetric group on six elements.
    It is a group of order 6!=720 with group operation permutation composition.
    Here is an example: \theta =\left( {\begin{array}{cccccc}<br />
   1 & 2 & 3 & 4 & 5 & 6  \\<br />
   2 & 3 & 1 & 4 & 6 & 5  \\ \end{array} } \right).

    So in this is standard notation 3,4,5,~\&~6 must be assigned.

    Does this differ with the definition in your textbook?
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  5. #5
    Member elninio's Avatar
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    It seems to be the second case, where there could be a different amount of members that satisfy theta.
    I'm confused as to how I would account for the possibility of it including 3,4,5 and all 6 members.
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  6. #6
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    I have already answered this question.
    I don’t understand your confusion unless you text material differs from a standard.
    I asked you if it did, but you choose not to answer.

    Here it is again.
    There are 4! permutations in S_6 each of which assigns 1\to 2~\&~2\to 3.

    What is your problem?
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  7. #7
    Member elninio's Avatar
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    My apologies. I thought your answer implied that all and only 6 elements were part of that cycle.
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