Presumably we are going to regard as identical colourings of the cuboid which can be obtained from each other by rotation.

If the cuboid is \(x\times y \times z\) in dimension, then it has three pairs of different faces, the \(x\times y\) faces, the \(x \times z\) faces and the \(y \times z\) faces.

Suppose that one of the colours we are using is red. There are \(3\) choices as to which type of face will be painted red. Once we have made that choice, rotate the cuboid until the red face is underneath. There are \(5\) choices for the colour of the topmost face (the other face of the same type as the one painted red).

Consider one of the four remaining colours. There are two choices as to which type of face will be painted that colour. Once we have made that choice, rotate the cuboid so that the red face is still underneath, and our newly painted face is facing the front. We cannot rotate the cuboid any more, so there are \(3!\) ways of painting the remaining three faces.

This gives us a total of \(3 \times 5 \times 2 \times 3! = \frac{6!}{4}\) colourings.

Another way of seeing this is the following: There are \({6 \choose 2}\) choices for the pair of colours to paint the \(x\times y\) faces, then \({4 \choose 2}\) further choices for the colours with which to paint the \(x \times z\) faces, with the remaining two colours going with the last pair of faces. There are just two ways of painting the cuboid with these colour pairs (think why!), so the number of colourings is
\[ {6 \choose 2} \times {4 \choose 2} \times 2 \; = \; \frac{6!}{4} \]

in my opinion because the all ways is 6! and we count each one 8 times the solution is 6!/8 (changing the color of not adjacent faces) but some of my friends say that the solution is 6!/4 can you help me.

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TopNewestPresumably we are going to regard as identical colourings of the cuboid which can be obtained from each other by rotation.

If the cuboid is \(x\times y \times z\) in dimension, then it has three pairs of different faces, the \(x\times y\) faces, the \(x \times z\) faces and the \(y \times z\) faces.

Suppose that one of the colours we are using is red. There are \(3\) choices as to which type of face will be painted red. Once we have made that choice, rotate the cuboid until the red face is underneath. There are \(5\) choices for the colour of the topmost face (the other face of the same type as the one painted red).

Consider one of the four remaining colours. There are two choices as to which type of face will be painted that colour. Once we have made that choice, rotate the cuboid so that the red face is still underneath, and our newly painted face is facing the front. We cannot rotate the cuboid any more, so there are \(3!\) ways of painting the remaining three faces.

This gives us a total of \(3 \times 5 \times 2 \times 3! = \frac{6!}{4}\) colourings.

Another way of seeing this is the following: There are \({6 \choose 2}\) choices for the pair of colours to paint the \(x\times y\) faces, then \({4 \choose 2}\) further choices for the colours with which to paint the \(x \times z\) faces, with the remaining two colours going with the last pair of faces. There are just two ways of painting the cuboid with these colour pairs (think why!), so the number of colourings is \[ {6 \choose 2} \times {4 \choose 2} \times 2 \; = \; \frac{6!}{4} \]

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in my opinion because the all ways is 6! and we count each one 8 times the solution is 6!/8 (changing the color of not adjacent faces) but some of my friends say that the solution is 6!/4 can you help me.

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6! / 4

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