# Number of Permutations for L5C?



## 2180161 (Oct 21, 2015)

Two questions.
1. title (Last 5 corners)
2. would it be suitable for speedsolving? (probably not)

What I think it is: 5!*3^4/15. Is this correct?

EDIT: 
What would it be if a certain corner were in a certain spot everytime? i.e Setup R U R' U' The FDR corner is always in URF. How many cases for that?


----------



## joshsailscga (Oct 21, 2015)

You'll have to elaborate on this a bit...

Edit: Also, check out the 'new methods' thread, there are plenty of people there who might be able to help you out.


----------



## 2180161 (Oct 21, 2015)

joshsailscga said:


> You'll have to elaborate on this a bit...
> 
> Edit: Also, check out the 'new methods' thread, there are plenty of people there who might be able to help you out.



The corners in the U layer are all unsolved. FDR is also scrambled.

That is what I mean by L5C (Last 5 corners) 

Setup: U' L' U2 R U' R' U2 L U'

How many permutations for a case like this (corners only) where RFD is in UFR


----------



## RhysC (Oct 21, 2015)

Personally I'd have something like this.

(4!*3^5)/12


----------



## 2180161 (Oct 21, 2015)

RhysC said:


> Personally I'd have something like this.
> 
> (4!*3^5)/12



How did you get that?


----------



## leeo (Oct 21, 2015)

There are 1009 corner 3-cycles from any buffer. If I multiply all of these by each other and filter out only the pure 5-cycles and not count duplicates, I get 108'864 total corner 5-cycles.


----------



## AlphaSheep (Oct 21, 2015)

I get 5!/2 possible permutations (must be even parity as edges are solved) and 3^4 possible orientations (orientation of the 5th corner is set by the other four). That gives 4860 cases. Having a particular corner in a specific location reduces it by a factor of 5.


----------



## RhysC (Oct 21, 2015)

2180161 said:


> How did you get that?



Sorry made a mistake. The four is meant to be a five.

(5!*3^5)/12

5! = number of permutations for 5 corners
3^5 = number of orientations for 5 corners
/12 = dividing all parities

In a similar fashion,

(12!*2^12*8!*3^8)/12 = 43 quintillion


----------



## shadowslice e (Oct 21, 2015)

RhysC said:


> Sorry made a mistake. The four is meant to be a five.
> 
> (5!*3^5)/12
> 
> ...



I'm slightly guessing here but wouldn't it be (5!*3^4)/12 becuase the last corner orientation depends on the previous 4

So there would be (if there was 1 fixed corner) (4!*3^4)/12= 168 possibilities

In the same way, ((12!*2^11)*(8!*3^7))/2 =4.32*10^16=~43 quintillion

I divided by 2 as there has to be an even number of 2-swaps, the 12!*2^11 denotes the number of edge positions and the 8!*3^7 denotes the corners.


----------



## RhysC (Oct 21, 2015)

shadowslice e said:


> I'm slightly guessing here but wouldn't it be (5!*3^4)/12 becuase the last corner orientation depends on the previous 4
> 
> So there would be (if there was 1 fixed corner) (4!*3^4)/12= 168 possibilities



Ugh got it wrong again, must be drunk or something today lol
(5!*3^5)/12 = 2430 possible ab5c
This video probably explains it better: https://www.youtube.com/watch?v=QV9k6dRQQe4

Fixed corners don't matter since we're dividing by all the parities.


----------



## RhysC (Oct 21, 2015)

WAIT!

Forgot flipped edges don't matter! 
(5!*3^5)/6 = 4860

My final verdict!


----------



## shadowslice e (Oct 21, 2015)

yeah I get 4860 too now I've had a look over it though I went with (5!*3^4)/2
(cominations*possible orientations)/even permutation

Or (4!*3^4)/2 = 972 with one solved corner.


----------



## collppllzbf2lll (Oct 21, 2015)

I think it's 602, because:
CLL, TCLL+ and TCLL-, this is for 1 corner in the FDR spot, and there are 5 corners so 40x3x5=600 + 2 pll algs?
if you're including AUF's for every top layer case it would be a lot more i think...
correct me if i'm wrong.


----------



## RhysC (Oct 21, 2015)

collppllzbf2lll said:


> I think it's 602, because:
> CLL, TCLL+ and TCLL-, this is for 1 corner in the FDR spot, and there are 5 corners so 40x3x5=600 + 2 pll algs?
> if you're including AUF's for every top layer case it would be a lot more i think...
> correct me if i'm wrong.



I don't think mixing alg sets is a great idea for working out combos, especially when they're for 2x2, not 3x3


----------



## shadowslice e (Oct 21, 2015)

collppllzbf2lll said:


> I think it's 602, because:
> CLL, TCLL+ and TCLL-, this is for 1 corner in the FDR spot, and there are 5 corners so 40x3x5=600 + 2 pll algs?
> if you're including AUF's for every top layer case it would be a lot more i think...
> correct me if i'm wrong.



I think the lower number is because you can AUF in TCLL but not this.


----------



## tseitsei (Oct 21, 2015)

Without auf-ing it is easy to figure out that it is:

5*4*3 = 5!/2 For permutations because 1st corner can be in any of the 5 spots 2nd in any of the remaining 4 and 3rd in the remaining 3 spots. The order of last to pieces is already known here because all edges are solved. So it becomes 5*4*3*1*1 actually.

3^4 For orientations because 4 out of 5 corners can be in any of the 3 orientations so 3*3*3*3 and once again the last corner must be in the only orientation that makes cube solvable so that's just 3*3*3*3*1.

Put those together and you have (5!/2)*3^4


----------



## 2180161 (Oct 21, 2015)

Sorry, I forgot to mention that this is for 2x2/ or with edges solved


----------



## shadowslice e (Oct 21, 2015)

2180161 said:


> Sorry, I forgot to mention that this is for 2x2/ or with edges solved



As I said above those give pretty different numbers.

Q


tseitsei said:


> Without auf-ing it is easy to figure out that it is:
> 
> 5*4*3 = 5!/2 For permutations because 1st corner can be in any of the 5 spots 2nd in any of the remaining 4 and 3rd in the remaining 3 spots. The order of last to pieces is already known here because all edges are solved. So it becomes 5*4*3*1*1 actually.
> 
> ...



Yep this is pretty much the exact same logic I went through with my calculation.


----------

