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Permutations of a group

  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    How many distinct permutations are there of the form (abc)(efg)(h) in S7?
    2. Relevant equations
    3. The attempt at a solution

    since we have 7 elements I think for the first part it should be 7 choose 3 then 4 choose 3.
    And then we multiply those together.
     
  2. jcsd
  3. Feb 8, 2016 #2

    Samy_A

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    Is the permutation (abc)(efg)(h) different from the permutation (efg)(abc)(h)?
     
  4. Feb 8, 2016 #3
    no, so I then need to divide by 3!
     
  5. Feb 8, 2016 #4

    haruspex

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    How do you arrive at 3?
     
  6. Feb 8, 2016 #5
    i mean 3 factorial, because (abc)(efg)(h)= (efg)(abc)(h) so I need to divide by 6 because their are 6 ways to arrange that and we would be over counting.
     
  7. Feb 8, 2016 #6

    haruspex

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    Yes, I understood it was factorial, but I was specifically challenging the 3.
    I only see two equivalent arrangements there. What other four have you redundantly counted by your method?
     
  8. Feb 9, 2016 #7
    oh your saying only divide by 2! because by the time we choose our last element h we only have one choice, so it wont affect the counting.
    Because by the time (abc)(efg) have been chosen we only have one choice left.
     
  9. Feb 9, 2016 #8

    haruspex

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    Yes, 2, but not for that reason.
    You chose abc as a set of three, so you covered all permutations of those three in one selection - no double counting so far.
    Likewise in choosing the set def.
    The double counting arises because you could have chosen the set def first, then the set abc.
     
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