Permutation combinations that will be possible

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My math is outdated now , but the formula will always help to identify this issue I am facing .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

[phillip1882]

i'm afraid your question is too poorly worded to really give an adequate answer.

the best i can do is try to explain what i THINK you're saying and then answer that question.

what you have is a pathing question, and you want to know how many different paths you can take.

lets take a few examples and see if we can come up with a general formula.

its always good to start with the dead obvious cases, to make things clear.

A 1 path

A B 2 paths (A, B)

A->(1) B 3 paths (A, A:1, B)

A->(1,2) B->(1,2,3) 7 paths (A, A:1, A:2, B, B:1, B:2, B:3)

A->(1,2) A:1->(1,2,3) A:2->(1,2) B->(1,2,3,4) B:2->(1,2) B:3->(1,2,3) 18 paths

(A, A:1, A:2, A:1:1, A:1:2, A:1:3, A:2:1, A:2:2, B, B:1, B:2, B:2:1, B:2:2, B:3, B:3:1, B:3:2, B:3:3, B:4)

is this correct?

Edited November 11, 2012 by phillip1882

[mathy_math]

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

Edited November 12, 2012 by mathy_math

[mathy_math]

Hello all

 

 

 

 

was my description clear the second time around ?

 

 

Thanks for any insight here on this path query.

 

 

[phillip1882]

if the number of paths at each split is fixed, then the general equation is...

r^(n+1) -r

------------- +1

r -1

 

if not then I'm afraid you'll just have to add.

 

lets say number of paths at each split is 5 and you have 4 levels.

5^(4+1)-5

-------------- + 1 = 781

5-1

Edited November 21, 2012 by phillip1882

[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

 

 

 

Edited November 21, 2012 by mathy_math