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# 3-color cube

## Question

If each side of a tetrahedron is an equilateral triangle painted white or black, five distinct color patterns are possible: (1) all sides white, (2) all black, (3) just one side white and the rest black, (4) just one side black and the rest white, and (5) two sides white, while the other two are black.

If each side of a cube is painted red or blue or yellow, how many distinct color patterns are possible?

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I got

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57 distinct color patterns. Still double checking my work, though...

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2 R'S  1B)  BRR ;RBR ; RRB = 3      1R& 2B ) RBB; BRB ;BBR .=3    2R& 1 Y ) YRR ;RYR; RRY;   =3    2Y & 1 R )  RYY ; YRY ; YYR ;=3                                                                                                 2B & 1 Y )    YBB ;  BYB ' BBY ;  = 3       2Y& 1B )   BYY ;  YBY ;YYB ;=3         RY&B )   RYB ; RBY ; YRB ; YBR ; BYR ; BRY ;= 6

TOTAL 6 X 3 = 18  PLUS 6 = 24

Edited by bonanova
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2R'  1B)  BRR ;RBR ; RRB = 3      1R& 2B ) RBB; BRB ;BBR .=3    2R& 1 Y ) YRR ;RYR; RRY;   =3    2Y & 1 R )  RYY ; YRY ; YYR ;=3                                                                                                 2B & 1 Y )    YBB ;  BYB ' BBY ;  = 3       2Y& 1B )   BYY ;  YBY ;YYB ;=3         RY&B )   RYB ; RBY ; YRB ; YBR ; BYR ; BRY ;= 6

TOTAL 6 X 3 = 18  PLUS 6 = 24      CORRECTION : in the 1st listed BRR; assuming you are looking at B front and back, then the middle face  upward is either R or Y;                      the same can be said for the R  i.e. the upward face must be B or Y; The 2nd R  upward is either B or Y .So these possibilities exist  reading from left to right                                    R B B;    R B Y ;  R Y B ; R Y Y     YBB ; YBY ; YYB ; YYY  = 8  .    Each cell or horizontal possibility  has a total of 8 poss due to the center face of the cube.                                      Since 1 possible  has been counted in the original calculation then 8- 1 =  7.      The grand total is 24 x 7  = 168 possibilities

Edited by bonanova
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I think...

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1 color: 3 possibilities

2 colors:

a. one side a different color from the others: 6 possibilities

b. two sides a different color: 6 choices of colors and 2 configurations adds 12 possibilities

c. three sides a different color than the other three: 3 choices of colors, 2 configurations, 6 more possibilities

3 colors:

a. four of one color, one each of other colors: 3 ways to choose colors, 2 configurations, 6 possibilities

b. three of one color, two of another, one of another: 6 ways to choose colors, 3 configurations, 18 possibilities

c. two of each color: 1 color choice, 4 configurations, 4 more possibilities

...so, in total, there should be 3 + (6 + 12 + 6) + (6 + 18 + 4) = 55 possible cubes

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On 12/15/2017 at 1:43 AM, rocdocmac said:

If each side of a tetrahedron is an equilateral triangle painted white or black, five distinct color patterns are possible: (1) all sides white, (2) all black, (3) just one side white and the rest black, (4) just one side black and the rest white, and (5) two sides white, while the other two are black.

If each side of a cube is painted red or blue or yellow, how many distinct color patterns are possible?

I'll just point out the interesting fact that with a tetrahedron, saying "two sides white" exhausts all permutations, since all pairs of sides share an edge. But with the cube this is not the case. So, I'm wondering what "patterns" is intended to mean: combinations? or permutations?

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5 hours ago, bonanova said:

I'll just point out the interesting fact that with a tetrahedron, saying "two sides white" exhausts all permutations, since all pairs of sides share an edge. But with the cube this is not the case. So, I'm wondering what "patterns" is intended to mean: combinations? or permutations?

This is a good point.  I should clarify that for my answer, I assumed that if one configuration of colors can be transformed into another by rotating the cube, then they should not be counted as two distinct configurations.

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20 hours ago, ThunderCloud said:

I think...

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1 color: 3 possibilities

2 colors:

a. one side a different color from the others: 6 possibilities

b. two sides a different color: 6 choices of colors and 2 configurations adds 12 possibilities

c. three sides a different color than the other three: 3 choices of colors, 2 configurations, 6 more possibilities

3 colors:

a. four of one color, one each of other colors: 3 ways to choose colors, 2 configurations, 6 possibilities

b. three of one color, two of another, one of another: 6 ways to choose colors, 3 configurations, 18 possibilities

c. two of each color: 1 color choice, 4 configurations, 4 more possibilities

...so, in total, there should be 3 + (6 + 12 + 6) + (6 + 18 + 4) = 55 possible cubes

Oops, I missed a couple... a slight correction:

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...

3 colors:

c. two of each color, 6 configurations, 6 more possibilities...

In total, 3 + (6 + 12 + 6) + (6 + 18 + 6) = 57 possible cubes. I think Pickett got it.

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Colored cube.xlsx

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1 hour ago, rocdocmac said:
Reveal hidden contents

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I think the answer is really 57. There are two cubes missing from the spreadsheet... it is possible for there to be exactly two sides of each color such that each side is adjacent to another side of the same color. There are two distinct ways to do this, poorly drawn here:

Edited by ThunderCloud
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Oooops, really slipped up here! Fifty seven appears to be the correct answer. Attached the updated solution  with the two addtional arrangements added to the bottom row on the right.

My sincere apology to all participants.

Colored cube.xlsx

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