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Homework Help: Permutation List

  1. Feb 29, 2012 #1
    Hi to everyone....

    I would like to know which is the permutation 4661 of the elements G = {1,2,3,4,5,6,7}
    I have search for the formula around the internet but I could not find it...
    So I hope any one can help about this solution.
    Thank you...

    and the result on the picture is like this as you can see, as I mention I cant find any formula that can figure this example...

    Attached Files:

  2. jcsd
  3. Mar 1, 2012 #2
    You can think about it like this:
    How many permutations will start with 7? What range of positions will they occupy?
    Of those, how many will have 3 second?

    You won't get a formula, as such, but there will be an algorithm or process for producing the correct sequence.
  4. Mar 1, 2012 #3
    Hello Joffan, I'll try to explain again a bit better because my English in math way isn't that good, so I'll try to explain as best as I can...
    So 7! Factorial is equal to 5040 permutation in the list and that permutation is 7,6,5,4,3,2,1
    but What I need is the permutation 4661 in the list of this numbers ???
    And my work so far is like this :

    Number one 1 appears to be 720 times in the first position it doesn't change 720 times!!!
    1>>>> 1,2,3,4,5,6,7
    2>>>> 1,2,3,4,5,7,6
    3>>>> 1,2,3,4,6,5,7
    4>>>> 1,2,3,4,6,7,5
    5>>>> 1,2,3,4,7,5,6
    6>>>> 1,2,3,4,7,6,5
    7>>>> 1,2,3,5,4,6,7
    8>>>> 1,2,3,5,4,7,6
    9>>>> 1,2,3,5,6,4,7
    720>>>> 1,7,6,5,4,3,2
    In the row 121, and here is where it changes and where number 2 gets in the first position for along next 720 times which is equal to 1440... How ?

    721>>>> 2,1,3,4,5,6,7
    722>>>> 2,1,3,4,5,7,6
    723>>>> 2,1,3,4,6,5,7
    724>>>> 2,1,3,4,6,7,5
    1440>>>> 2,7,6,5,4,3,1
    And I have gather this numbers for (1)720, for 2(720), for 3(720), for 4(720), for 5(720)and for 6(720) ====>> 720+720+720+720+720+720 = 4320, and the permutation of 4320 is >>>>> 6,7,5,4,3,2,1....... I didn't include number 7 in here because automatically we get 5040 permutation's which 7,6,5,4,3,2,1.
    So what I need, is to find 4661 permutation on the list ???

    So I hope I have explain in the good way...
    and I hope you can help me with this example...ASAP
    Thank you
  5. Mar 1, 2012 #4
    I understood the problem - I just wanted you to be involved in the exploration of the problem.

    Ok, so you get that the first member changes on blocks of 720 (=6!). In the same way, the second member changes on blocks of 5!, the third on blocks of 4!, 3!, 2!, 1! and 0! (=1, conventionally).

    So we want to look at which batch of 720 we're in, then within that which batch of 120, then which batch of 24, etc.

    Thinking about that first choice carefully, it's almost like 1+floor(N/720), but that just misses on the boundary - N=720 should actually gives us a 1 for first postion. And the same applies at every batch boundary, so we should should assess N-1 in place of N.

    We can actually do this more conveniently from the other end... which batch of 1 are we in (trivial), then which batch of 1 within two, then which batch of 2 within 6, then which batch of 6 within 24, etc.

    Then we get to the administrative business of not picking the same value twice - but that's not too hard either.
  6. Mar 6, 2012 #5
    Joffan thank you for help, I guess I solved out your way, this permutation but I have another one so i'll publish this as soon as possible so the other can understand how it works...
  7. Mar 12, 2012 #6
    Solved !!!
    I guess this example will help you to find your permutation...
    Thank you.

    Attached Files:

  8. Mar 12, 2012 #7
    Good work, nicely laid out.

    If you want to waste a few days, try investigating the ways that the full permutation list can be generated by swapping one pair of adjacent elements between successive entries. :-)
  9. Mar 21, 2012 #8
    Hi Joffan, I am sorry for late reply but I was a kind a busy working and studying but I think I know what you want me to do and I will do it... I wish you wonderful day and a best of luck :D
  10. Mar 21, 2012 #9
    If you're busy, toli, please don't bother - it was only a suggestion for something else to explore in this general field.

    Mapping the permutations which differ by swapping one adjacent pair produces a graph - a 6-regular graph in this case - and the question becomes one of finding a Hamiltonian path through that graph.
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