Simple and Tricky

Assuming that we shouldn't count cubes that could be rotated to be the same as other cubes:

* Apply any color to the top: 1 choice
* Apply a different color to the bottom: 5 choices
* Apply any color to the front: 1 choice (any color can be rotated to the front)
* Apply a color to the left: 3 choices
* Apply a color to the back: 2 choices
* Apply a color to the right: 1 choice.

Answer: 5*3*2 = 30 :roll:

6!
=720

In how many ways can a cube be painted using 6 different colors such that all the sides of the cube have a different color?

A) 720
B) 120
C) 60
D) 30
E) 24

Let the 6 colors = ABCDEF.

The orientation of the cube is irrelevant.
We need to count the number of ways that the colors can be arranged RELATIVE to each other.
Let's work with one pair of colors -- A and B -- and calculate the number of ways that the 4 other colors can be arranged RELATIVE to A and B.
There are 2 options: A and B are on ADJACENT faces, or A and B are on OPPOSITE faces.

Case 1: A and B on ADJACENT faces.
Number of ways to arrange the 4 other colors = 4! = 24.

Case 2: A and B on OPPOSITE faces.
Let A = the base of the cube and B = the top of the cube.
The 4 remaining colors 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 remaining colors = (4-1)! = 3! = 6.

Total number of ways = 24+6 = 30.

The correct answer is D.

GMATGuruNY wrote:

In how many ways can a cube be painted using 6 different colors such that all the sides of the cube have a different color?

A) 720
B) 120
C) 60
D) 30
E) 24

Let the 6 colors = ABCDEF.

The orientation of the cube is irrelevant.
We need to count the number of ways that the colors can be arranged RELATIVE to each other.
Let's work with one pair of colors -- A and B -- and calculate the number of ways that the 4 other colors can be arranged RELATIVE to A and B.
There are 2 options: A and B are on ADJACENT faces, or A and B are on OPPOSITE faces.

Case 1: A and B on ADJACENT faces.
Number of ways to arrange the 4 other colors = 4! = 24.

Case 2: A and B on OPPOSITE faces.
Let A = the base of the cube and B = the top of the cube.
The 4 remaining colors 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 remaining colors = (4-1)! = 3! = 6.

Total number of ways = 24+6 = 30.

The correct answer is D.

Hi GMATguruNY,

I did not understand this...Sorry i could not point to one step as i am not clear from the beginning. Can you please explain it bit more clearly...I was thinking the answer would be 720 as 6!..But i am really surprised to see this...Why we need to consider one adjacent condition and the opposite side condition...Hope i am clear in my question...Thank you....

@GMATGuruNY:
Could your please tell how the concept of the question is different from the following question's:
There are 6 chairs and 6 different people. in how many different ways they can sit?

Here they did not specify about any arrangement. We have 6 colors and 6 faces.
1st face can be painted with any of 6 colors.
2nd face can be painted with any of remaining 5 colors.
3rd face can be painted with any of remaining 4 colors.
......and so on.

So the number of ways a cube can be painted with diff colors is: 6*5*4*3*2*1=720.

Please let me know if this approach is wrong.

nandy1984 wrote:

GMATGuruNY wrote:

In how many ways can a cube be painted using 6 different colors such that all the sides of the cube have a different color?

A) 720
B) 120
C) 60
D) 30
E) 24

Let the 6 colors = ABCDEF.

The orientation of the cube is irrelevant.
We need to count the number of ways that the colors can be arranged RELATIVE to each other.
Let's work with one pair of colors -- A and B -- and calculate the number of ways that the 4 other colors can be arranged RELATIVE to A and B.
There are 2 options: A and B are on ADJACENT faces, or A and B are on OPPOSITE faces.

Case 1: A and B on ADJACENT faces.
Number of ways to arrange the 4 other colors = 4! = 24.

Case 2: A and B on OPPOSITE faces.
Let A = the base of the cube and B = the top of the cube.
The 4 remaining colors 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 remaining colors = (4-1)! = 3! = 6.

Total number of ways = 24+6 = 30.

The correct answer is D.

Hi GMATguruNY,

I did not understand this...Sorry i could not point to one step as i am not clear from the beginning. Can you please explain it bit more clearly...I was thinking the answer would be 720 as 6!..But i am really surprised to see this...Why we need to consider one adjacent condition and the opposite side condition...Hope i am clear in my question...Thank you....

Unlike a row of seats, a CUBE can be INVERTED and ROTATED.

Thus, the following represent only ONE way to arrange A and B:
A on the bottom face and B on the top face.
B on the bottom face and A on the top face.
A on the front face and B on the back face.
B on the front face and A on the back face.
A on the left face and B on the right face.
B on the left face and A on the right face.

In each of these arrangements, A and B are on opposite faces.
In each arrangement, the cube could be INVERTED and/or ROTATED so that A is on the bottom and B is on the top.
Thus, painting A and B on ANY PAIR of opposite faces yields only ONE possible arrangement of A and B.

The same holds true if A and B are painted on adjacent faces: in each case, the cube could be INVERTED and/or ROTATED so that A is on the bottom and B is on the front.
Thus, each PAIR of adjacent faces yields only ONE POSSIBLE WAY to paint A and B on the cube.

My solution above ensures that no duplicate arrangements will be counted.
When the cube is painted, in EVERY POSSIBLE arrangement of the 6 colors, there are ONLY TWO OPTIONS for A and B: either A and B are on ADJACENT faces or A and B are on OPPOSITE faces.

If A and B are on ADJACENT faces, the number of ways to arrange the 4 other colors = 4! = 24.

If A and B are on OPPOSITE faces, the number of ways to arrange the 4 other colors in a CIRCLE = (4-1)! = 6.

Thus, the total number of ways to arrange the 6 colors on the cube = 24+6 = 30.

shankar.ashwin wrote:In how many ways can a cube be painted using 6 different colors such that all the sides of the cube have a different color?

A) 720
B) 120
C) 60
D) 30
E) 24

This question tests circular permutation. Do we get such questions in GMAT?BTW the solution --

total ways - 6! = 720

But, the cube has 13 axes and rotating a cube along those axis will give you 24 identical cubes.

hence we have to divide the count (6!) by 24

ANSWER -- 720/24 = 30 ways...

EDIT: - I know that we do get questions were people are arranged around the circular table etc. Cube is bit more complicated case. Luckily, I remember the concept