TITLE: What would be the number of inequivalent $6$-colourings of the faces of a cube?
QUESTION [7 upvotes]: Consider the different ways to colour a cube with $6$ given colours such that each face will be given a single colour and all the six colours will be used. Define two such colourings to be equivalent if one will get from another just by rotation. Then what would be the number of inequivalent colourings?

The answer is given $30$. But I was trying to prove it by choosing the colour in the two opposite sides. 
For example, first let us choose the colours on the two opposite sides in $^6C_2$ ways, then another two opposite sides in $^4C_2$ ways, and the leftover sides in $^2C_2$ ways. So, in total, we get $15 \times 6=90$, but I think I am triple-counting the situation or thinking it in a wrong way.
Please help.

REPLY [6 votes]: In a circular permutation, arrangements which may be obtained from one another by rotating the cube are considered identical.
Place a colour on the bottom.  It does not matter which.  The top colour may be chosen in five ways.  Place another colour facing you.  Again, it does not matter which.  The remaining three colours may be arranged relative to the colour facing you in $3!$ ways as we proceed clockwise around the cube.  Hence, there are $5 \cdot 3! = 30$ admissible colourings.