Possible Permutations & Combinations

[hello_math]

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My math is outdated now , but the formula will always help to identify this issue I am facing .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

 

[tnich]

Your links are broken, but I think I understand what you mean.
Suppose you were to design a computer program to compute the number of combinations. (I am not suggesting that you actually do that, although it wouldn't be very hard, just that you think about how you would design a computer program to compute the answer to your problem.) If I understand what you are trying to do, I think you could design the program as a set of nested loops. If you can do that, then I think you can write out the equation for the answer.

 

[blackmarch103]

Thanks you for sharing.