Problem Solving

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Problem Solving

by kavn » Wed Feb 01, 2012 9:59 am
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)

(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

OA:E

The actual answer is B, but I am not sure how to get to that.
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by GMATGuruNY » Wed Feb 01, 2012 12:58 pm
kavn wrote:You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)

(A) 24
(B) 30
(C) 48
(D) 60
(E) 120

OA:E

The actual answer is B, but I am not sure how to get to that.
Let the 6 colors = ABCDEF.
Let's work with one pair of colors, A and B.
There are only 2 options: A and B on ADJACENT faces, or A and B on OPPOSITE faces.

Case 1: A and B on ADJACENT faces.
The number of ways to place A and B on adjacent faces = 1.
Why? Because if A and B are on ANY TWO ADJACENT FACES, the cube can be positioned so that A is on the bottom and B is in front, yielding only ONE WAY to place A and B on adjacent faces.

Now we need to count the number of ways to arrange C, D, E and F RELATIVE to A and B.
The number of ways to arrange the 4 colors C, D, E and F = 4! = 24.

To combine our options for A and B with our options for C, D, E and F, we mulitply:
1*24 = 24.

Case 2: A and B on OPPOSITE faces.
The number of ways to place A and B on opposite faces = 1.
Why? Because if A and B are on ANY TWO OPPOSITE FACES, the cube can be positioned so that A is on the bottom and B is on top, yielding only ONE WAY to place A and B on opposite faces.

Now we need to count the number of ways to arrange C, D, E, and F RELATIVE to A and B.
If A is on the bottom and B is on top, C, D, E and F need to be arranged on the 4 remaining faces: the front face, the left face, the back face, and the right face.
These 4 faces form a circle around the middle of the cube.
The number of ways to arrange N elements in a circle = (N-1)!.
Thus, the number of ways to arrange the 4 colors C, D, E and F in a circle = (4-1)! = 3! = 6.

To combine our options for A and B with our options for C, D, E and F, we mulitply:
1*6 = 6.

Total ways:
Number of ways in Case 1 + number of ways in Case 2 = 24+6 = 30.

The correct answer is B.

Please note that this problem seems WAY beyond the scope of the GMAT: I've never seen an actual GMAT question about the rotation of a cube.
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