# Possible Permutations & combinations

by hello_math
Tags: combinations, permutations
 P: 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/a...23_01/13-2.jpg Here's a hand sketch of what I have been trying to explain above , for A and B respectively. http://i856.photobucket.com/albums/a...23_01/13-1.jpg 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 My math is outdated now , but the formula will always help to identify this issue I am facing .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

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