Problem Solving

abhi332 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

[spoiler]OA:B[/spoiler]

Total number of ways to paint the cube = (total number of ways to arrange the 6 colors)/(total number of ways to orient the cube).

Total number of ways to arrange the 6 colors = 6! = 720.
Total number of ways to orient the cube = 24.
Total number of ways to paint the cube = 720/24 = 30.

The correct answer is B.

Here is one way to count the number of ways that each arrangement can be oriented:
Let say that the 6 colors are A, B, C, D, E and F and that the cube has been painted so that A and B are on opposite faces.
The cube can be oriented so that any of the 6 colors is on top.
Number of options for the color on the top face = 6.
Let's say that the cube has been oriented so that A is on top.
Since A and B are on opposite faces, B is on the bottom.
The cube can now be rotated clockwise so that either C, D, E or F is facing forward.
Number of options for the forward face = 4.
To combine the number of options for the top face and the number of options for the forward face, we multiply:
6*4 = 24.

(D) 60

Number of ways = (6*5*4*3*2*1)/((3*2*1)*2)

This problem has reference to Circular Permutation with flipping allowed, i.e. (1/2)*(6-1)! or 60 possible ways.

arunspanda wrote:This problem has reference to Circular Permutation with flipping allowed, i.e. (1/2)*(6-1)! or 60 possible ways.

The Answer cannot be 60..
6! --> Number of ways of arranging color..
4! --> the number of ways 4 - side sided walls can interchange ..

6!/4! = 6x5 = 30

The easiest piece of this problem is determining the number of ways to paint the top of the cube, which is obviously 6.

For the second half of this problem we need to use the formula for a combination of n objects taken r at a time: nCr = n!/[r!(n-r)!]to determine how many unique combinations we can paint the four visible sides of the cube.

Specifically, we want to use the formula for combinations because we are only interested in counting one specific pattern of colors on the sides (red, yellow, blue and green AND green, red, yellow and blue should only be counted as one distinct pattern, for instance).

The tricky part is realizing that we have five remaining colors to choose from for the four remaining visible sides of the cube (the color of the bottom side has no bearing on our calculations for this particular problem).

So far, we know we have six options for the color of the top of the cube, and we have five options for the colors of each of the four remaining visible sides of the cube. For the sides, we're going to use our formula above to find the number of combinations of 5 objects taken 4 at a time, which gives us 5C4 = 5!/[4!(1!)] = 5.

Since we want to know the number of distinct ways of combining the possible choices for the top AND sides of the cube, we multiply 6 x 5 = 30, or choice (B).

abhi332 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

[spoiler]OA:B[/spoiler]

The steps to solve the question are as follows:

1) There has to be some base on which the cube is resting which can be colored by any one color as reorienting the cube doesn't make any difference. So let's assume that one color and one face is fixed for base of Cube leaving us with 5 colors and 5 faces of cube to be painted

2) let's select one color for the top face now which can be selected in 5 ways

3) Remaining four faces are forming like a circle which can be painted by 4 colors in 3! ways

Now total ways of painting the faces of cube = 5 x 3! = 5 x 6 = 30

Answer: Option B

I disagree with everyone so far!
If you like solving a Rubik cube, you might like this solution:


We will start with any corner of a cube.
Three faces are adjacent to this corner.
CLOCKWISE about this corner:
Using colours 1, 2 and 3, there are exactly 2 unique colour arrangements: 123123123... and 132132132...
Using colours 1, 2 and 4, there are exactly 2 unique colour arrangements: 124124124... and 142142142...
Using colours 1,2 and 5, there are exactly 2 unique colour arrangements: 125125125... and 152152152...
Using colours 1,2 and 6, there are exactly 2 unique colour arrangements: 126126126... and 162162162...
In other words, the first corner has 2 X 4 = 8 unique colour arrangements.

For the opposite corner, there are only 3 colours (A, B and C) remaining to choose from.
Using colours A, B and C, there are exactly 2 unique colour arrangements: ABCABCABC... and ACBACBACB...
HOWEVER, by rotation they can be oriented relative to the first corner, 3 different ways (which is allowed, as the previous statement demonstrates that ABCABCABC is uniquely different to ACBACBACB).
Therefore the second corner has 2 X 3 = 6 unique colour arrangements.

Therefore, combining the possibilities for both diametrically opposed corners, there are 8 x 6 = 48 unique arrangements. So I propose that the answer is [C]! Please discuss.

GMATGuruNY wrote:

abhi332 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

[spoiler]OA:B[/spoiler]

Total number of ways to paint the cube = (total number of ways to arrange the 6 colors)/(total number of ways to orient the cube).

Total number of ways to arrange the 6 colors = 6! = 720.
Total number of ways to orient the cube = 24.
Total number of ways to paint the cube = 720/24 = 30.

The correct answer is B.

Here is one way to count the number of ways that each arrangement can be oriented:
Let say that the 6 colors are A, B, C, D, E and F and that the cube has been painted so that A and B are on opposite faces.
The cube can be oriented so that any of the 6 colors is on top.
Number of options for the color on the top face = 6.
Let's say that the cube has been oriented so that A is on top.
Since A and B are on opposite faces, B is on the bottom.
The cube can now be rotated clockwise so that either C, D, E or F is facing forward.
Number of options for the forward face = 4.
To combine the number of options for the top face and the number of options for the forward face, we multiply:
6*4 = 24.

I tend to disagree. If the top face is chosen, there are only 5 choices for the bottom face.
Then out of the 4 faces remaining, one must be fixed (as its 4 different orientations amount to the same thing), leaving only 3 choices.
This would suggest 3 x 5 arrangements = 15.
This even differs from my previous answer of 48!
It seems that there are many methods that produce sensibly different answers.

GMATGuruNY wrote:

abhi332 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

[spoiler]OA:B[/spoiler]

Total number of ways to paint the cube = (total number of ways to arrange the 6 colors)/(total number of ways to orient the cube).

Total number of ways to arrange the 6 colors = 6! = 720.
Total number of ways to orient the cube = 24.
Total number of ways to paint the cube = 720/24 = 30.

The correct answer is B.

Here is one way to count the number of ways that each arrangement can be oriented:
Let say that the 6 colors are A, B, C, D, E and F and that the cube has been painted so that A and B are on opposite faces.
The cube can be oriented so that any of the 6 colors is on top.
Number of options for the color on the top face = 6.
Let's say that the cube has been oriented so that A is on top.
Since A and B are on opposite faces, B is on the bottom.
The cube can now be rotated clockwise so that either C, D, E or F is facing forward.
Number of options for the forward face = 4.
To combine the number of options for the top face and the number of options for the forward face, we multiply:
6*4 = 24.

I tend to disagree. If the top face is chosen, there are only 5 choices for the bottom face.
Then out of the 4 faces remaining, one must be fixed (as its 4 different orientations amount to the same thing), leaving only 3 choices.
This would suggest 3 x 5 arrangements = 15.
This even differs from my previous answer of 48!
It seems that there are many methods that produce sensibly different answers.

Mathsbuddy wrote:
I tend to disagree. If the top face is chosen, there are only 5 choices for the bottom face.
Then out of the 4 faces remaining, one must be fixed (as its 4 different orientations amount to the same thing), leaving only 3 choices.
This would suggest 3 x 5 arrangements = 15.
This even differs from my previous answer of 48!
It seems that there are many methods that produce sensibly different answers.

"Disagreeing" is the fundamental right that we all have got but let me clarify the mistake...

The mistake is highlighted by the bold part

There are four faces which are to be painted will serve like a circle with 4 beads and if you have to arrange 4 beads then you fix one of them which you did but the remaining three faces have to be pained by 3 colors which can happen in 3 x 2 x 1 way (3 possibility for first of remaining 3 faces, 2 for the remaining two faces as one has been pained out of 3 unfixed and 1 way to paint last face)

Therefore the four faces can be painted by 3! ways and hence the answer becomes 5 x 3! = 5 x 6 = 30

I hope it explains it all... If it doesn't then relax for 5 mins, take a small walk and then read the explanation once again and think again. You will eventually find it correct.

[It isn't funny... This "relaxing for 5 mins" really works well]

emalby wrote:taken from another website:

Let the six faces be a, b, c, d, e, and f. With face a opposite face b there are six arrangements for the other four colours around the cube: cdef, cdfe, cedf, cefd, cfde and cfed. Likewise for the face a opposite face c ; face a opposite face d ; face a opposite face e ; and face a opposite face f. All have six arrangements for the remaining five colours. Hence the total is 5 x 6 = 30 arrangements.

I would agree with this solution. Please ignore my earlier rantings!

Thanks GMATinsight.
It's all clear to me now. I knew that my "disagreements" would be incorrect, as I concurred with 3 different answers! I'm glad you didn't think me rude. It was just a debate to explore the truth.
I am convinced now the answer is 30. Thank you.

GMATinsight wrote:

Mathsbuddy wrote:
I tend to disagree. If the top face is chosen, there are only 5 choices for the bottom face.
Then out of the 4 faces remaining, one must be fixed (as its 4 different orientations amount to the same thing), leaving only 3 choices.
This would suggest 3 x 5 arrangements = 15.
This even differs from my previous answer of 48!
It seems that there are many methods that produce sensibly different answers.

"Disagreeing" is the fundamental right that we all have got but let me clarify the mistake...

The mistake is highlighted by the bold part

There are four faces which are to be painted will serve like a circle with 4 beads and if you have to arrange 4 beads then you fix one of them which you did but the remaining three faces have to be pained by 3 colors which can happen in 3 x 2 x 1 way (3 possibility for first of remaining 3 faces, 2 for the remaining two faces as one has been pained out of 3 unfixed and 1 way to paint last face)

Therefore the four faces can be painted by 3! ways and hence the answer becomes 5 x 3! = 5 x 6 = 30

I hope it explains it all... If it doesn't then relax for 5 mins, take a small walk and then read the explanation once again and think again. You will eventually find it correct.

[It isn't funny... This "relaxing for 5 mins" really works well]

IMO:B