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In how many ways can 6 faces of a rectangular parallelopiped with all 3 dimensions distinct , be painted with 6 different colours??

I have tried and i am getting 90 by $\displaystyle\frac{6!}{2^3}$. Thankyou

Lehs
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hiten
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  • Are you saying that each face must be a different color? Or just that there are 6 available colors, and some could be used more than once (and others not at all)? – paw88789 Oct 13 '14 at 10:57
  • I think the first condition that each face must be a different color – hiten Oct 13 '14 at 11:14

1 Answers1

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Let's call the faces (in opposite pairs) left, right; top, bottom; and front, back. Pick two colors for left, right. At this point it doesn't matter which of these two colors goes on which face since left, right can trade places by turning the object. So this selection can be done in ${6\choose 2}=15$ ways.

Now do the same for front, back. Again which color goes on which of front, back doesn't matter, because we can turn the figure from one configuration to the other by rotating $180^\circ$ through the left-right axis. So this step can be done in ${4\choose 2}=6$ ways.

There are now two colors left for top and bottom, but now which face gets which color matters (you can't rotate one configuration into the other with the other 4 faces already painted). So there are 2 ways to do this step.

So the total number of colorings is $15\cdot 6\cdot 2 = 180$.

@hiten: I think in your solution you divided by one too high a power of $2$ because of the distinguishability of switching the last two opposite faces being colored.

paw88789
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    Good. Alternatively, you can use Burnside's lemma. The symmetry group of the parallelepiped has order $4$; there are $6!=720$ colorings; because of the condition that each face is a different color, none of the colorings is invariant under any of those symmetries except the identity; so the number of distinguishable colorings is $720/4=180$. – bof Oct 13 '14 at 12:30