mathy_math

Ok Thanks a Lot ! I shalll give this one a try and should get a fair indication of whats going on here. I have a rough sketch of the manual paths I have taken and I can tally it with this formula. Thanks you replied on some other forums , I did not get any response at all

Hello all was my description clear the second time around ? Thanks for any insight here on this path query.

I am sorry for the confusion due to the wordings and explanations. But yes, you are right. I have indeed a pathing question, and want to know how many different paths can be taken here. [ this is the hand drawing which I am referring to : http://i856.photobuc...123_01/13-1.jpg ] 'A' will always have 1 Path [ so : A --->Path to # 1---->then a path to # 1_4 ] is 1 Path of the many that are possible 'B' too will always have 1 Path [ similar to the above example , but then for B ] -Yes you are correct : A B 2 paths (A, B) - A will ALWAYS travel through 1 , 2 & 3 [ further the # 1 can have *_1 till *_5 values ] / [ #2 can have * _1 till *_3 values] & [ # 3 can have *_1 till *_6 values ] In your notation (math notation )- (For A) : A, A:1, A:2, A:3 ,A:1:1, A:1:2, A:1:3, A:1:4, A:1:5, A:2:1, A:2:2, A:2:3 , A:3:1 , A:3:2, A:3:3, A:3:4, A:3:5, A:3:6 this logic holds for B too B, B:4, B:5, B:6 ,B:4:1, B:4:2, B:4:3, B:4:4, B:4:5, B:5:1, B:5:2, B:5:3 , B:6:1 , B:6:2, B:6:3, B:6:4, B:6:5, B:6:6 and finally we will have A + B I hope the description was a bit clear this time ? Hope to hear back . Thanks

I'm called Shovon and I like maths. so joined this forum .

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 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 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