# Combinations on other puzzles



## Joel Banks (May 4, 2016)

Let's find the combinations on some of the other, less talked about puzzles. Explain how you found the combinations or probabilities on a puzzle. I'll start it off, how many combinations are on a megaminx? How many more times is this that a Rubik's cube?
(Other ideas, pyraminx, gigaminx, master pyraminx, cuboids?)


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## Daniel Lin (May 4, 2016)

megaminx: 3^19*2^29*20!/2*30!/2=too big for my calculator
each corner can have 3 orientations and there are 20 corners. Last corner is fixed. Each edge can have 2 orientations and there are 30 edges. Last edge is fixed. There are 20!/2 ways to permute the corners and 30!/2 ways to permute edges. You divide by two because odd parity is not possible on megaminx.


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## Joel Banks (May 4, 2016)

Awesome thanks!


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## YouCubing (May 4, 2016)

Daniel Lin said:


> megaminx: 3^19*2^29*20!/2*30!/2=too big for my calculator
> each corner can have 3 orientations and there are 20 corners. Last corner is fixed. Each edge can have 2 orientations and there are 30 edges. Last edge is fixed. There are 20!/2 ways to permute the corners and 30!/2 ways to permute edges. You divide by two because odd parity is not possible on megaminx.


Googled it, apparently it's a little more than 10^68 a.k.a. one hundred unvigintillion.


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## Daniel Lin (May 4, 2016)

Someone do skewb, I'm interested


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## AlphaSheep (May 4, 2016)

You may be interested in http://www.jaapsch.net/puzzles/

You can see the number of possible combinations for more puzzles than you even thought existed.


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## guysensei1 (May 4, 2016)

I'll ask one here, how many positions are there on the Pentultimate?

Note that there are multiple orientations that one position can be in.


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## bcube (May 4, 2016)

guysensei1 said:


> how many positions are there on the Pentultimate?



Third google´s result: http://www.twistypuzzles.com/~sandy/forum/viewtopic.php?f=1&t=9262&start=0


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## gamesgamerton55 (Mar 5, 2021)

How about gigaminx or teraminx


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## gamesgamerton55 (Mar 5, 2021)

How?


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## Christopher Mowla (Mar 6, 2021)

gamesgamerton55 said:


> How about gigaminx or teraminx


The number of positions for the minx^2 to minx^20. (minx^3=normal size, minx^5 = gigaminx, minx^7 = teraminx, etc.)

See this post for my derivation of this expression for a minx of size _n_ (where _n_ can be *even* too!).



gamesgamerton55 said:


> How?


I'm not sure either why he said 6. If it's not one position, then it should be 24. (But I personally only think there is one position for the 1x1x1.)


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## Gnome (Mar 6, 2021)

Christopher Mowla said:


> But I personally only think there is one position for the 1x1x1.



Agreed, 1 position (or state), 24 orientations (or external state changes that do not affect the internal state).


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## Melvintnh327 (Aug 28, 2021)

why does the 4x4 have 7.4 quattuordecillion possible combinations? no matter how I calculate it, I just couldn't get that number.

Edit: never mind I figured it out.

\[ \frac{8!\times3^7\times24!\times24!}{24\times4!^6}=7.4011968416\times10^{45} \]


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## xyzzy (Aug 28, 2021)

Melvintnh327 said:


> quattuordecillion


Protip: avoid using the names for powers of 10 greater than 10^12, and definitely don't use them above than 10^33. "Quadrillion" (10^15) up to "decillion" (10^33) might still be understandable with some basic knowledge of Latin numbers, but it gets increasingly esoteric thereafter. (Not to mention that short scale is just _completely messed up_. Long scale made sense; why did we have to switch away from it…)

Anyway, yeah, your calculation is right, if you treat rotationally equivalent states as being the same. You can get the number by fixing a corner piece in place, by fixing an edge piece in place, or by dividing out the number of rotationally equivalent states (which happens to be a constant 24, _because_ you can fix a corner piece or an edge piece). On a puzzle where every piece type has duplicates somewhere (e.g. some stickermods), it's possible that the numbers of rotationally equivalent states (the sizes of the equivalence classes, in other words) are no longer constant and you can't just divide by 24.


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## Melvintnh327 (Aug 28, 2021)

xyzzy said:


> Protip: avoid using the names for powers of 10 greater than 10^12, and definitely don't use them above than 10^33. "Quadrillion" (10^15) up to "decillion" (10^33) might still be understandable with some basic knowledge of Latin numbers, but it gets increasingly esoteric thereafter. (Not to mention that short scale is just _completely messed up_. Long scale made sense; why did we have to switch away from it…)


*cough cough* I just like these names because I like math... *cough cough*
like for example: Unquadragintaquadringentillion ( \( 10^{1326} \) )


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## abunickabhi (Aug 29, 2021)

Melvintnh327 said:


> *cough cough* I just like these names because I like math... *cough cough*
> like for example: Unquadragintaquadringentillion ( \( 10^{1326} \) )


Thats an insanely big number. Its still much smaller than Graham's number though.


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## Melvintnh327 (Aug 29, 2021)

abunickabhi said:


> Thats an insanely big number. Its still much smaller than Graham's number though.


That number that I wrote is actually 6 times smaller than the number of possible combinations of the 19x19 rubik's cube


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## Christopher Mowla (Aug 29, 2021)

Melvintnh327 said:


> That number that I wrote is actually 6 times smaller than the number of possible combinations of the 19x19 rubik's cube


No offense, but this thread is about combinations on *other* puzzles. So continuously bring up the nxnxn is off-topic. There are plenty of threads on the combinations of the nxnxn.

And this thread is certainly not about fascination with weird names given to large integers.

Thanks!


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## CubeRed (Aug 30, 2021)

How about combinations...
On a fisher cube?


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## Megaminx lover (Aug 30, 2021)

1x1x1 pyraminx?


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## CubeRed (Aug 30, 2021)

Megaminx lover said:


> 1x1x1 pyraminx?


Good question. Now I will attempt to answer the question all the elite brains in MIT have failed to answer. Trying to calculate the sheer amount of 0 there is in that number will result in a self maintenance system embedded to the codes in the Internet to override resulting various nuclear weapons to undergo a destruction program to the G7 countries. But I have spent several hours working 5 Pentagon level quantum computers to work it out. Behold the equation...

3x3x3x3=12

Of course, this is assuming you don't count the rotations as a variable.If you included that it would result in a USA military aircraft containing a biological weapon to unleash in Russia resulting superhuman beings with superior genome sequencing and nuclear mutations to take over Earth. I hope you don't do that...


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## Megaminx lover (Aug 30, 2021)

3x3x3x3 is 81.


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## CubeRed (Aug 30, 2021)

Megaminx lover said:


> 3x3x3x3 is 81.


Yeah LOL
For da meme


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## Megaminx lover (Aug 31, 2021)

CubeRed said:


> Yeah LOL
> For da meme


?


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## CubeRed (Aug 31, 2021)

Megaminx lover said:


> ?


*Cough cough*
Anyway... So is anyone doing the fisher cube combination?


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## Matt11111 (Aug 31, 2021)

CubeRed said:


> *Cough cough*
> Anyway... So is anyone doing the fisher cube combination?


Off the top of my head, 4 of the centers have 4 possible orientations, so that multiplies the number of possible permutations by 1024.

The middle layer edges look the same in both of their orientations, so it's possible to flip an even number of them and the cube will still look identical. I believe that means that for every permutation, there are 7 others that look the same with different edge orientations, so divide by 8. So if I'm not mistaken that adds up to 5,536,256,419,134,701,568,000 permutations on a Fisher cube


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