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I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.

So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.

And for finding the number of permutations in x amount of edges and x amount of corners, I believe you just multiply both of the number of permutations together?

Thanks.

And Note: This is finding the number of solvable permutations of course, e.g., not finding permutations of flipped pieces as well.

And also, I should have clarified, This is for 3x3.

I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.

So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.

And for finding the number of permutations in x amount of edges and x amount of corners, I believe you just multiply both of the number of permutations together?

Thanks.

And Note: This is finding the number of solvable permutations of course, e.g., not finding permutations of flipped pieces as well.

I am designing an app that calculates the number of permutations in a certain amount of edges, corners, or edges corners combined.

So it would be greatly appreciated if someone could tell me how to calculate the number of permutations in x amount of edges, and how to calculate the number of permutations in x amount of corners. This would be GREATLY appreciated.

And for finding the number of permutations in x amount of edges and x amount of corners, I believe you just multiply both of the number of permutations together?

Thanks.

And Note: This is finding the number of solvable permutations of course, e.g., not finding permutations of flipped pieces as well.

And also, I should have clarified, This is for 3x3.

12! - ways to permute edges
2^12 - ways to orient edges
8! - ways to permute corners
3^8 - ways to orient corners

However, we have to divide by:

2 - since the last edge's orientation is determined by the previous 11
3 - since the last corner's orientation is determined by the previous 7
2 - since the permutation parity of the puzzle is always even (i.e. can't swap only 2 corners)

This gives:

12! * 2^(12-1) * 8! * 3^(8-1) / 2

So if you want to do this for a subset of x edges and y corners, fixing the rest, this function could be used to find the number of permutations of those pieces:

f(x, y) = x! * 2^(x-1) * y! * 3^(y - 1) / 2

Note that this is not the same as cases for LL, as counting specific cases takes symmetries and AUF into account. So if you're looking for literal permutations this will work, but it gets trickier if your purpose is for counting specific cases for a given alg set.

Terrific! Doing a large simulation to get a more reliable estimate (about 18 million random perms) I get a chance of 1 in 833 for a cycle 1260. So the estimate would be 4.325e19/833 = 5.19e16. Comparing to your 5.15e16, I'd call bingo within the MOE. So the answer is 51,490,480,088,678,400. A...

I don't think there's really a way to figure this out, but for some reason a lot of the misscrambles in comps result in really fast times like Max's OH solve at Nats. If someone who isn't really world-class has a misscramble, technically nobody would know about it since no one would notice an average solve. So that chances that someone like Max or Felik's 1: gets a misscramble cubed, and 2 solves it in WR or extremely close to it would be astronomical right? But it has happened multiple times, so I maybe wrong scrambles happen more often than we think?

So that chances that someone like Max or Felik's 1: gets a misscramble cubed, and 2 solves it in WR or extremely close to it would be astronomical right?

While 1 is moderately unlikely in a well-run competition, 2 isn't all that unlikely for those two people! There's also some reason to believe that misscrambles caused by misreading a single move in a scramble sequence (as opposed to misscrambles caused by accidental move done in transit, misscrambles with 2 or more misread moves, etc.) are slightly biased towards being easier than typical.

Introduce enough fudge factors and you can get something that sounds like it should be unlikely, but in reality is very likely. This is something you have to be very wary of when doing statistics.

Yeah, this is the classic blade of grass paradox. In a pre-COVID world, world class solvers were each often doing upwards of 100 solves across all events every month. Additionally, there are hundreds of world class cubers whose solves are closely followed by the community. So it may seem like a one-in-a-million occurrence to happen at that very point in time, but with the frequency that official solves happen at a high level, it's almost certain for it to happen in a way that is noticed eventually.