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how many times can i...

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Chaoslord
also the thing with the cubes can be confusing, i normaly would say 6! but, if you say fix one point, and chose the next and so on then you could get a rotation of a cube which has already been accounted for

think about it like this, pick a colour x and chose the other sides, if all the 4 of the sides change could and only 2 remain constant they cube is not a new cube it is a rotation on a used one.

What you're talking about here is slightly different to the basic nPr stuff, because you've added an extra rule about rotating the cube...

So think about it this way:
You have six colours. Paint one side of the cube with any colour.
Next, you have 5 colours to choose from to paint the face of the cube that is OPPOSITE the one you just painted.
Paint another side with any remaining colour.
You now have 3 colours to choose from to paint the OPPOSITE face.
Paint another side with any remaining colour.
You now have 1 colour to choose from (not much of a choice, though...) to paint the last face.

So your answer would be: 5x3x1=15 different combinations of colours of the cube, that cannot be rotated to look the same.
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Chaoslord
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tomthecool
What you're talking about here is slightly different to the basic nPr stuff, because you've added an extra rule about rotating the cube...

So think about it this way:
You have six colours. Paint one side of the cube with any colour.
Next, you have 5 colours to choose from to paint the face of the cube that is OPPOSITE the one you just painted.
Paint another side with any remaining colour.
You now have 3 colours to choose from to paint the OPPOSITE face.
Paint another side with any remaining colour.
You now have 1 colour to choose from (not much of a choice, though...) to paint the last face.

So your answer would be: 5x3x1=15 different combinations of colours of the cube, that cannot be rotated to look the same.


i agree but im sure how you derived you logic =/ im sure its correct but how can you prove there is no repartition?
Chaoslord
i agree but im sure how you derived you logic =/ im sure its correct but how can you prove there is no repartition?

I don't know about "proving" it, but just think about it...

No matter how you rotate the cube, two opposite sides will remain in the same (opposite) positions relative to each other. So these are the only pairs of sides that you are interested in.

Actually, this is exactly the sort of question you'd be likely to get in the maths challenge. You just need to think about the problem carefully and find the easy solution.

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