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So, do they asked this on test day? This drove me nuts.

Imagine a flattened cube... The three colored region will establish the other colors of the remaining faces of the cube.

For example: We assumed the sequence of color in the given image as RED on face#1 and BLUE on face#2 and GREEN on face#3. Since face#1 is RED then we know #4 and #5 cannot be RED. Since face #2 is BLUE, we know that #5 and #6 cannot be BLUE. Since face#3 is GREEN, we know #4 and #6 (the bottom) cannot be GREEN.

So, all we need is to count the possible number of arrangements of 3 colors.

3! = 6
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A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three colors, red, blue, and green are used to paint the six faces of the cube. If the adjacent faces are painted with the different colors, in how many ways can the cube be painted?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 27

If the base of the cube is red, then in order the adjacent faces to be painted with the different colors, the top must also be red. 4 side faces can be painted in Green-Blue-Green-Blue OR Blue-Green-Blue-Green (2 options).

But we can have the base painted in either of the three colors, thus the total number of ways to paint the cube is 3*2=6.

Answer: B.

Dear Bunuel
With the top and base of the cube being red, whether 4 side faces are Green-Blue-Green-Blue OR Blue-Green-Blue-Green (2 options) it should not matter because using the top and base as the axis, we can rotate Green-Blue-Green-Blue[i] to be [i]Blue-Green-Blue-Green right?

So according to the solution, why there are 2 distinct options?
Those 2 options should be counted as 1 because it can be rotated.

Please shed some light
Thank you!
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A cube marked 1, 2, 3, 4, 5, and 6 on its six faces. Three colors, red, blue, and green are used to paint the six faces of the cube. If the adjacent faces are painted with the different colors, in how many ways can the cube be painted?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 27

If the base of the cube is red, then in order the adjacent faces to be painted with the different colors, the top must also be red. 4 side faces can be painted in Green-Blue-Green-Blue OR Blue-Green-Blue-Green (2 options).

But we can have the base painted in either of the three colors, thus the total number of ways to paint the cube is 3*2=6.

Answer: B.

Dear Bunuel
With the top and base of the cube being red, whether 4 side faces are Green-Blue-Green-Blue OR Blue-Green-Blue-Green (2 options) it should not matter because using the top and base as the axis, we can rotate Green-Blue-Green-Blue[i] to be [i]Blue-Green-Blue-Green right?

So according to the solution, why there are 2 distinct options?
Those 2 options should be counted as 1 because it can be rotated.

Please shed some light
Thank you!

The point is that the faces are numbered. Let me rephrase the solution. Say face #1 is painted red, then the opposite face (say it's numbered 6) should also be red. Faces, 2, 3, 4, and 5, should be Green-Blue-Green-Blue OR Blue-Green-Blue-Green. There is a difference between these two options because Green=2, Blue=3, Green=4, Blue=5 is different from Blue=2, Green=3, Blue=4, Green=5.
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Dear Bunuel VeritasKarishma EMPOWERgmatRichC

I have one follow-up question from the reply above:

If the same question is asked but this time the cube is NOT marked, there will be only ONE way in which we can paint the cube right?
(The way is that each of the 2 opposite sides is of the same color. And however we paint it, we could rotate the cube so that it is still the same coloring.)

Thank you in advance!
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Hi varotkorn,

YES - if there was nothing distinguishing the 6 sides of the cube (in the original prompt, the numbers distinguish the sides) AND we were asked to paint with only 3 colors AND if no two adjacent sides could be the same color, then there would be just ONE way to paint the cube. The 'orientation' of the cube would not be a factor, since any of the sides could the top/bottom, left/right and front/back sides. However, the moment the individual 6 sides DO become relevant, then the answer would change.

GMAT assassins aren't born, they're made,
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