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Possible Permutations & combinations

  1. Nov 9, 2012 #1
    Hi all

    I am new , and wanted to ask the following. I do not know if this is the right section but here goes:

    I have three registers , say A , B and C

    Case 1:A will always have 3 combinations 1,2,3

    1 has further subsections 1_1,1_2,1_3,1_4,1_5
    2 has further subsections 2_1,2_2,2_3
    3 has further subsections 3_1,3_2,3_3,3_4,3_5,3_6

    A will always have to go through 1, 2 ,3 and will have 1 path of any subsection

    Example , a possible combination

    A--(will always traverse)1--(and will end with one of the subs)1_1
    |
    |
    |--(will always traverse)2--(and will end with one of the subs)2_2
    |
    |
    |--(will always traverse)3--(and will end with one of the subs)3_4


    another example

    A--1--1_5
    |
    |
    |--2--2_1
    |
    |
    |--3--3_1

    ++++++++++++++++++++++++++++++++++++++++++++++++++++

    the same holds good for B too

    Case 2:B will always have 3 combinations 4,5,6

    4 has further subsections 4_1,4_2,4_3,4_4,4_5
    5 has further subsections 5_1,5_2,5_3
    6 has further subsections 6_1,6_2,6_3,6_4,6_5,6_6

    (Just like A) B will always have to go through 4, 5 ,6 and will have 1 path of any subsection

    Example , a possible combination

    B--4--4_3
    |
    |
    |--5--5_3
    |
    |
    |--6--6_4


    +++++++++++++++++++++++++++++++++++++++++++++++++++

    and Finally , there's a combination of (A+B)

    Case 3:
    - where A will again take the same path/s as mentioned for A above

    - where B will again take the same path/s as mentioned for B above

    Example , a possible combination

    A--1--1_5 B--4--4_3
    | |
    | |
    |--2--2_1 And |--5--5_3
    | |
    | |
    |--3--3_1 |--6--6_4

    if the above did not come all right here in the forum(my dabbings with the notepad , so here's the picture I want to show for Case 3:
    http://i856.photobucket.com/albums/ab124/Hello_123_01/13-2.jpg [Broken]

    Here's a hand sketch of what I have been trying to explain above , for A and B respectively.

    http://i856.photobucket.com/albums/ab124/Hello_123_01/13-1.jpg [Broken]
    Case 3 is as mentioned a (case1 +Case 2)



    My question is: How many combinations do I have , till I have exploited all permutations/combinations ? so all combinations/permutations that could be covered by Case1 , Case 2 and Case 3
    and what formula did you use to deduce it ?
    My math is outdated now , but the formula will always help to identify this issue I am facing .


    Thanks for any help here


    *PS: I have to mention, that the path is always linear. so for example:
    1 Path = B + 4+ 4_1 | B + 5+ 5_1 | B+6+6_1 = Linear path , Right Path
    next combination = B + 4+ 4_1 | B + 5+ 5_2 | B+6+6_1
    next combination = B + 4+ 4_1 | B + 5+ 5_3 | B+6+6_1
    next combination = B + 4+ 4_2 | B + 5+ 5_1| B+6+6_1


    and so on.....

    Path = B + 4+ 4_1 | B + 5+ 5_1/5_2/5_3 | B+6+6_1/6_2/6_3 = wrong , not this way (no simultaneous or multiple paths)

    I hope I was able to explain myself
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
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