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Magnus Eisengrim
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This product is a 3x3 tray of cubes. In the write-up the manufacturer claims that "billions of fascinating images to be created (68,719,476,736 possible combinations)." The number they give happens to be 2^36.

This seems wrong to me. Assuming each face is distinct and has no rotational symmetry and that all 9 cubes are in play, each cube can be placed in the tray in 24 ways. (6 faces x 4 orientations). The 9 cubes can be arranged in 9! ways. If I am right, we have 24x9!=8,709,120 permutations.

I find combinatorial problems difficult, so I'm sure I've made a basic error somewhere. But I have a hard time believing that 2^36 is right either.

Thanks

John
The blood-dimmed tide is loosed, and everywhere
The ceremony of innocence is drowned;
The best lack all conviction, while the worst
Are full of passionate intensity.--Yeats
Magnus Eisengrim
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On sober second thought, it seems that it should that be 24^9.

Geez.
The blood-dimmed tide is loosed, and everywhere
The ceremony of innocence is drowned;
The best lack all conviction, while the worst
Are full of passionate intensity.--Yeats
Magnus Eisengrim
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Sorry for thinking out loud on a public forum. My second post is just stupid. Sober second though? Ha!
The blood-dimmed tide is loosed, and everywhere
The ceremony of innocence is drowned;
The best lack all conviction, while the worst
Are full of passionate intensity.--Yeats
S2000magician
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Quote:
On 2011-03-07 15:51, Magnus Eisengrim wrote:
On sober second thought, it seems that it should that be 24^9.

I think it's (9!) * (24^9) ÷ 4 = 239,664,780,049,121,000.

I think.

(Decide which cube goes in which position: 9!. Decide the orientation of each cube: 24^9. Remove duplicates obtainable by rotating the frame: 4.)
Magnus Eisengrim
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Thank you. That was one of the approaches that I talked myself out of last night. I don't know why, but this kind of problem leads me all sorts of self-deceit.

I couldn't convince myself that there were only 4 rotational duplicates with your method.

John
The blood-dimmed tide is loosed, and everywhere
The ceremony of innocence is drowned;
The best lack all conviction, while the worst
Are full of passionate intensity.--Yeats
S2000magician
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Quote:
On 2011-03-08 09:06, Magnus Eisengrim wrote:
I couldn't convince myself that there were only 4 rotational duplicates with your method.

If you allow that you could flip the thing over, there are eight.

Imagine a simpler objet d'art: only one cube. There are 6 * 4 = 24 orientations, but the same four rotational duplicates; hence, 24 ÷ 4 = 6 orientations. That is, you pick which face is on top.
Magnus Eisengrim
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Thank you. Somehow the problem seem simple now Smile

John
The blood-dimmed tide is loosed, and everywhere
The ceremony of innocence is drowned;
The best lack all conviction, while the worst
Are full of passionate intensity.--Yeats
landmark
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The other day I was trying to figure how many unique 4x4 sudoku answers there could be. Harder than it looks. Can't say I'm right.
landmark
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Quote:
On 2011-03-08 00:09, S2000magician wrote:
Quote:
On 2011-03-07 15:51, Magnus Eisengrim wrote:
On sober second thought, it seems that it should that be 24^9.

I think it's (9!) * (24^9) ÷ 4 = 239,664,780,049,121,000.

I think.

(Decide which cube goes in which position: 9!. Decide the orientation of each cube: 24^9. Remove duplicates obtainable by rotating the frame: 4.)

Could be wrong, but those nine cubes look identical to me, in which case you don't need the factor of 9! Also some of the faces appear to have point symmetry (i.e. 180 degrees) and 90 degree rotational symmetry so that the orientation of some cubes may be double counted for some situations in the above. I think though that it only becomes an issue for the middle cube where all the other sides are hidden.
S2000magician
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On 2011-03-19 15:46, landmark wrote:
Could be wrong, but those nine cubes look identical to me, in which case you don't need the factor of 9! Also some of the faces appear to have point symmetry (i.e. 180 degrees) and 90 degree rotational symmetry so that the orientation of some cubes may be double counted for some situations in the above. I think though that it only becomes an issue for the middle cube where all the other sides are hidden.

It isn't clear to me whether the cubes are identical, so I did the calculation as if they weren't and - as you point out - there is no symmetry to each cube. If the cubes are identical or there are symmetries to consider, the number of visually distinct patterns is reduced accordingly.
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