Hi all

I am new ,was passing by 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My math is outdated now , but the formula will always help to identify this issue I am facing .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