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## Question

You have six different colors to paint a cube. Each face of the cube should be painted. You can use every color to paint as many faces of the cube as you like (0, 1, 2, 3, 4, 5, 6); but you cannot use more than one color on a single face.

How many different painting patterns can be formed on this cube?

Note: If a pattern can be formed by rotating another pattern any number of times, these two patterns are not considered as different.

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I just wrote a quickie program to get

2226

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I'm not sure why but...

6^6/4^2 = 2916

6^6 = total number of colored cubes with no conditions

4^2 = the number of different (identical) rotations for each cube.

Although I learned not to use my instincts in probabilities.

But this feels like the ultimate battle between Emotions & Machine (Computer code)

And the winner is ... again.

Edited by roolstar
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just to make you laugh...

this is why you dont ask an artist this question.

I was thinking in actual colors, not patterns.

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I'm not sure why but...

6^6/4^2 = 2916

6^6 = total number of colored cubes with no conditions

4^2 = the number of different (identical) rotations for each cube.

Although I learned not to use my instincts in probabilities.

But this feels like the ultimate battle between Emotions & Machine (Computer code)

And the winner is ... again.

only 4^2 of different rotations for each cube but 4*6. So then solution would be 1944.

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only 4^2 of different rotations for each cube but 4*6. So then solution would be 1944.

I think your count is too small.

You are correct that there are 24 rigid-motion

symmetries of the cube. But, in dividing

66 by 24 you are tacitly assuming

that every coloring is counted 24 times.

This is not true for coloring the entire cube

with a single color, for example. Cubes which

are colored with 2 colors also have fewer than

24 symmetric brothers. So, to get a correct

count, you would have to break the problem up

into sets of colorings which have the same

number of symmetric copies. This seemed difficult

to me, which is why I wrote a program to brute

force the thing.

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419 904

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superprismatic is right. I didn't get an answer that agreed with any of the ones posted, so I searched online and found this nifty formula.

(n^6 + 3n^4 + 12n^3 + 8n^2) / 24 = number of coloring of a cube with n colors

So n=6 gives 2226 possible colorings.

Taken from the wikipedia page for Burnside's Lemma

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That's a great lemma. I just figured out how to use it using the following links:

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I think your count is too small.

You are correct that there are 24 rigid-motion

symmetries of the cube. But, in dividing

66 by 24 you are tacitly assuming

that every coloring is counted 24 times.

This is not true for coloring the entire cube

with a single color, for example. Cubes which

are colored with 2 colors also have fewer than

24 symmetric brothers. So, to get a correct

count, you would have to break the problem up

into sets of colorings which have the same

number of symmetric copies. This seemed difficult

to me, which is why I wrote a program to brute

force the thing.

I think I agree...

And the Brain wins again... What a shocker!

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it should be "any" instead of "every". confused me a little. with that being said, i still dont know what your talking about.

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Working from first principles, the first color can be put on any of the 6 faces, but due to rotational symmetry, this can be counted only as 1.

The second color can be put in 2 ways - either as one of the four adjoining faces (which are symmetrical), or on the opposite faces. The third color can be put in any of the remaining 4 positions (no more symmetries available now). The 4th color goes into 3 available spaces and the 5th color goes into one of the remaining 2 faces while the 6th color goes into the only face left. The total options are 1X2X4X3X2X1=48. Can anyone tell me if there is a fallacy in this argument?

Edited by pankajvarma
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Working from first principles, the first color can be put on any of the 6 faces, but due to rotational symmetry, this can be counted only as 1.

The second color can be put in 2 ways - either as one of the four adjoining faces (which are symmetrical), or on the opposite faces. The third color can be put in any of the remaining 4 positions (no more symmetries available now). The 4th color goes into 3 available spaces and the 5th color goes into one of the remaining 2 faces while the 6th color goes into the only face left. The total options are 1X2X4X3X2X1=48. Can anyone tell me if there is a fallacy in this argument?

The problem begins to be easily seen as you begin with the third color:

If the first two colors were on opposite sides of the cube,

then placing the third color can be done in only one way

as placing it on one of the four remaining faces is

symmetrical with placing it on any other. However, if

the first two colors were on adjacent faces then there

are several different ways to place the third: adjacent

to both and adjacent to one but not the other. The

placing of the fourth color is even more complicated.

So, using first principles, you must chase down all the

possible paths and build up the number of ways in this

manner. It gets complicated and is quite confusing.

That's why I chose to write a program to produce all

66 colorings, then I started rotating each

one to see if it is a duplicate of any other, throwing

it out if it was. The program was just replicating what

a human would want to do but couldn't because of the

enormous boring work.

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