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Proving that these are all possible permutations

  1. Feb 15, 2012 #1
    So if I have numbers 1,2,3, intuitively you can say that there are 6 permutations:


    If there any way to rigorously prove that these are the only possible arrangements of these three numbers? Even more simple, is there any way to prove that 12 and 21 are the only permutations of 1 and 2? I don't need a ton of detail if it's going to be tedious, I'm just curious.

  2. jcsd
  3. Feb 15, 2012 #2
    There are only 3! = 6 possible arrangements (three possibilities for the first position, two possibilities for the second, and one for the third), and you've listed six different arrangement. Thus, you listed all possible arrangements.
  4. Feb 15, 2012 #3


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    Hey maxbashi and welcome to the forums.

    There is actually a set of formal definitions for permutations that you could use and would be helpful if you had to do more general or abstract permutations but for this specific problem it might be wise to use Number Nines advice and use the combinatoric identity for the number of unrestricted permutations (i.e. 3! = 6).
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