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If you have a 6-sided cube that has a unique color on each side (6 different colors), how many possible combinations of unique cubes are there? What if we change the problem to say there are 6 colors to choose from but not all sides must be unique?

I can't seem to wrap my head around a good way to approach this problem. Whatever I seem to try I always end up thinking about a way to turn the dice that messes up my counting.

Thanks for any help.

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For the first one, unless I am missing something, it is $\frac{6!}{24} = \frac{\text{Total number of possibilities}}{\text{Number of symmetries}} = 30$. –  user17762 May 18 '12 at 1:39
    
Can you elaborate on how you managed to get the number of symmetries? –  Matt May 18 '12 at 1:41
    
Andre has described it in his answer. –  user17762 May 18 '12 at 1:43

2 Answers 2

up vote 4 down vote accepted

To get rid of duplicate counts, put the cube on a table, with colour $1$ down. (i) If colour $2$ is not on the "up" side, rotate the cube until colour $2$ is facing you. (ii) If colour $2$ is on the up side, then rotate the cube (keeping colour $1$ down) so that colour $3$ is facing you.

Counting the possibilities in (i) should now be straightforward, as should counting the possibilities in (ii).

Alternately, there are $5$ possibilities for the "up" side. For each possibility, rotate the cube so that the smallest numbered remaining colour is facing you. Now the rest of the counting is not difficult.

Added: There are more general group-theoretic approaches (Polya counting), but it does not seem reasonable to bring them out for this fairly "small" problem.

For situations in which colours may be duplicated, a similar approach will work. However, because the possible symmetries are more complicated, Polya counting becomes more reasonable. But one can also break things down into cases, by examining separately $1$ colour, $2$, $3$, $4$, $5$, and $6$.

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The number of unique cubes is

$\frac{\text{# of ways to assign colors to the integers 1-6}}{\text{# of ways to rotate the die}}$

The denominator is 24 — 6 possibilities for the "top" face × 4 possibilities for the "front" face (adjacent to the "top" face).

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