Painting parallelopiped with 6 different color

Question

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

Answer 1

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.