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NJJ
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Imagine you have four dice of different colours. Each face of each dice has a different symbol on it from A to X.

Taking into account

a) the position of each dice in the stack. I.e. The 4 positions in the stack
b) the number of dice i.e. The 4 dice
c) the value of each dice. I.e. The 6 faces on each dice
d) the orientation of each die. e.g. the 4 ways that a dice can be turned when flat.

How many different ways can the dice be stacked?

I've tried I can't do the math!
NJJ
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576?
Kamal
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As best as I can figure, it's 576 as you said.

The number of combinations, for the order of the 4 dice is 4! or 24. Each dice can be permuted through 6 different sides, with each side having 4 possible orientations. In other words, 24 x 6 x 4 = 576.
Nir Dahan
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Be careful guys,
some symbols like X or O have a 2 fold or 4 fold symmetry - so you will over count combinations.

you'd have to list down your symbols first decide how many have rotational symmetry and take this into account.

N.
airship
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If you're talking capital letters (and yes, I know you're not using 'Y' or 'Z' - bear with me):

2-fold (rotational) symmetry: HISZ
4-fold symmetry: OX (unless 'squished' as they usually are in a standard font, in which case they have 2-fold symmetry)

There are other letters that can get you into trouble, depending on how they are rendered. If they are nice and square, 'M' and 'W' can be mirror images, as can 'Z' and 'N'.

Small letters have their own set of problems.

But if you ignore all of these potential complications and treat is as simply a problem in combinatorics: the guys up there who said '576' are absolutely right.
'The central secret of conjuring is a manipulation of interest.' - Henry Hay
Nir Dahan
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With small letters, you also have to consider "p", becoming a "d" (same issue as the M W mentioned above)
I think martin gardner (who else) had a similar problem with 2 cubes and a calendar.
can't find it at the moment
Scott Cram
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Nir, you're probably thinking of the "Two-Cube Calendar" puzzle from Mathematical Circus (Chapter 15). That really wasn't a stacking problem, though. You did have to figure out what the hidden faces were, and take into account certain orientations to get it right, though.
Nir Dahan
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I just checked the book. It is not it...
there was another question with cubes and letters. It is certainly gardner. If you find something else, just tell us where it is Scott.
I have almost all of Gardner's recreational math works, I'll try to hunt this puzzle tonight... guess it means I won't sleep much ...

N
Scott Cram
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Another possibility is the 3-cube calendar puzzle from the "Stanford" chapter of "The Magic Numbers of Dr. Matrix". You have to figure out how to distribute letters on 3 cubes so that they can spell any month's first three letters (and it refers back to the puzzle I mentioned in my previous post).
Nir Dahan
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I think that is the one...
thanks scott.
NJJ
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The orientation of the symbols themselves is irrelevant since rotating the symbol will also rotate the face itself.
WilburrUK
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I think the answer is 7,962,624

4! combinations of the order of the dice = 24
for each of those, 6 possible "up faces" per die 6*6*6*6 (1296)
for each of those, 4 possible orientations per die 4*4*4*4 (256)

total no. of combos = 24 * 1296 * 256 = 7,962,624

of course, this assumes that each die is stacked exactly on top of, and exactly aligned (rotationally) with the one below. It also assumes that there are 4 possible orientations for the bottom die which are significant (ie it's rotation relative to the rest of the world), if that isn't important then the answer is a quarter if the number above (so 1,990,656)

As Nicholas says above, the symmetry is a red herring (I think) because the "Up Face" defines a unique set of "Side Faces/Winding order" - in other words if the die were the other way up, then the same side-faces would show but the clockwise order of those face would be reversed. I may be possible to contrive to design the symbols in such a way that this causes duplicates - maybe you could put MpdW around one die, which would read the same on the inverted die. That would make the other "up faces" on that die give indistinguishable sides halving the number of apparent combos (but only when that die was not on top).
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