# Thread: Cube and Colours

1. ## Cube and Colours

If I had a cube and six colours and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had n colours instead of 6?

2. Originally Posted by Aquafina
If I had a cube and six colours and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had n colours instead of 6?
6! (asssuming that a red 1 spot side is part of a way that is different to a red 2 spot side etc.)

n(n-1) .... (n-5) (asssuming that a red 1 spot side is part of a way that is different to a red 2 spot side etc.)

3. I *think* that we are supposed to assume two colorings are equivalent if the cube can be rotated so that one coloring coincides with the other. If that is the case, then Mr. Fantastic's answer is too high.

4. Originally Posted by awkward
I *think* that we are supposed to assume two colorings are equivalent if the cube can be rotated so that one coloring coincides with the other. If that is the case, then Mr. Fantastic's answer is too high.
Perhaps the OP will clarify. In the meantime, if you want to post your solution under that assumption, please feel free to.

5. OK, let's say two colorings are equivalent if the cube can be rotated so that they coincide. We would like to count the non-equivalent colorings.

Let's say one of the colors is red. Place the cube on your desktop and position the red face on top. There are then 5 ways to color the bottom face. For each of these, there are 4! / 4 ways to color the side faces if we consider the 4 rotations of the cube, leaving the top and bottom in place, to be equivalent. So there are 5 * 4! / 4 = 30 ways in all.

(There is another way to approach this problem, using generating functions and the Polya Enumeration Theorem, but that's probably more advanced than you want to read about.)

6. Originally Posted by mr fantastic
6! (asssuming that a red 1 spot side is part of a way that is different to a red 2 spot side etc.)

n(n-1) .... (n-5) (asssuming that a red 1 spot side is part of a way that is different to a red 2 spot side etc.)
Thanks, how did you get the second one? By doing:

nC6 * 6! ?

7. Originally Posted by Aquafina
Thanks, how did you get the second one? By doing:

nC6 * 6! ?
n choices for the first side, (n-1) choices for the second etc. Take the product. Viola.

8. Originally Posted by awkward
OK, let's say two colorings are equivalent if the cube can be rotated so that they coincide. We would like to count the non-equivalent colorings.

Let's say one of the colors is red. Place the cube on your desktop and position the red face on top. There are then 5 ways to color the bottom face. For each of these, there are 4! / 4 ways to color the side faces if we consider the 4 rotations of the cube, leaving the top and bottom in place, to be equivalent. So there are 5 * 4! / 4 = 30 ways in all.

(There is another way to approach this problem, using generating functions and the Polya Enumeration Theorem, but that's probably more advanced than you want to read about.)
Hi, how did you get 4!/4? If we colour one of the side faces, we have 4 colours, then 3 for the next etc. Understand that.

How do you know to divide by 4 for the rotation? Also, since this is equivalent to 3!, can the answer be worked out by taking (n-1)! for the n side faces, or is that just for when there are 4 side faces?

THanks

9. Originally Posted by Aquafina
Hi, how did you get 4!/4? If we colour one of the side faces, we have 4 colours, then 3 for the next etc. Understand that.

How do you know to divide by 4 for the rotation? Also, since this is equivalent to 3!, can the answer be worked out by taking (n-1)! for the n side faces, or is that just for when there are 4 side faces?

THanks
We divide by 4 to account for the rotation because there are 4 rotations of the cube if we keep the top face in place. If the sides were an n-gon with n possible colors then there would be n! / n = (n-1)! ways to color the sides.