As I remember, this question was asked by my Mathematics teacher to the whole class and we discussed this question "only" in a whole lecture.
A cube, $6$ non distinguishable faces, is given. All we need to tell is the number of ways in which its faces can be coloured with $6$ different colours.
$1$. faces are to be coloured... and not edges !!
$2$. A face must be coloured with exactly one colour.
$3$. All six colours are to be used, say Blue, Green, Red, Yellow, Orange and White. as in the Rubik's cube.
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2This doesn't seem to have anything to do with probability. – hmakholm left over Monica Feb 15 '14 at 14:05
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1Does rotating the cube in space count as a "different" coloring? – hmakholm left over Monica Feb 15 '14 at 14:06
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Yes, just tell the number of different colourings... – ABcDexter Feb 15 '14 at 14:08
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If one configuration can be rotated to obtain another, are they distinct or no (I assume no, but want to make sure)? Must all six colors be used? – MT_ Feb 15 '14 at 14:10
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Please read the question again... – ABcDexter Feb 15 '14 at 14:11
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@AbcDexter: WHAT does it have to do with probability? Try "nothing" instead of "everything". – hmakholm left over Monica Feb 15 '14 at 14:15
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@AbcDexter Your edited post says the faces are not distinguishable. What do you mean by that? – Loki Clock Feb 15 '14 at 14:15
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@AbcDexter, if you wanted people to help you and answer your question, it would help if you would answer their clarifications. (Also, if you would accept their criticism of your tag choice, because this is combinatorics, not probability) – MT_ Feb 15 '14 at 14:15
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4@AbcDexter The Fibonacci numbers have application to pineapples. A question about Fibonacci numbers will not be tagged pineapples. The tags indicate essential subject matter, not the incidental. It lets answerers know the required expertise. – Loki Clock Feb 15 '14 at 14:20
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Ok, "nothing" instead of "everything"... – ABcDexter Feb 15 '14 at 14:23
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@Henning thanks for the answer... – ABcDexter Feb 15 '14 at 14:24
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This question has nothing to do with probability and has a simple solution with combinatorics...see my answer for that – Hawk Feb 15 '14 at 14:24
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@hawk , Henning already gave the correct solution. Thanks. – ABcDexter Feb 15 '14 at 14:28
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@AbcDexter I know he gave a solution and a damn good one at that...but it doesn't hurt to provide another solution...afterall that is what this community is about – Hawk Feb 15 '14 at 14:30
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Ok, what's your approach ?! – ABcDexter Feb 15 '14 at 14:32
3 Answers
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One approach: First imagine that the cube is fixed in space and cannot rotate. Then there are clearly $6!$ ways to distribute colors on the faces. But once we allow the cube to rotate, we find that we have counted each combination many times, namely one for each way the cube can be oriented in space. Each color combination has $6\cdot 4$ possible orientations, namely 6 directions the black face can point in, times 4 ways to then rotate the cube around the axis that passes through the center of that face. So the number of combinations is $$ \frac{6!}{4\cdot 6} = 5\times 3\times 2 = 30 $$
As another approach, we can divide into two cases: Either the black and the white face are neighbors, or they are opposite each other. If they are neighbors, we can choose to orient the cube with the black face up and the white face towards us, which completely specifies its orientation. Then the remaining 4 colors can be distributed in $4!$ ways.
If the black and white face are opposite, then orient the cube with the black face up, white face down and red face towards us. Then there remaining 3 colors can be distributed in $3!$ ways. So the number of combinations is
$$ 4! + 3! = 24 + 6 = 30 $$
hmakholm left over Monica
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I still don't know, how my teacher convinced me that the solution is exactly 1. – ABcDexter Feb 15 '14 at 14:21
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Call the six colors $1, 2, 3, 4, 5, 6.$ Put the cube on the table so that face $1$ is at the bottom. Consider face $2.$ If it is at the top then we can rotate the cube about a vertical axis so that face 3 is in front. Now the cube is fixed. There are $3!=6$ ways to complete the coloring. Now, suppose that face $2$ is a neighbor of $1.$ The we rotate the cube so that $2$ is in front. Now the cube is fixed, and the coloring can be completed in $4!=24$ ways. Altogether, there are $6+24=30$ distinct colorings of the cube by six colors.
Hawk
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There are $6!$ ways to color a cube, but we get many overcountings, and we want distinct colourings, so there are $24$ ways in which we can orient a cube. So, $\dfrac{6!}{24}=30$ ways of coloring the cube distinctly.
Hawk
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