Many years ago I set the following puzzle
in Hot Key: In how many ways can you colour the faces of a cube
using 6 colours? The colourings must be such that each coloured
cube must be unique whatever way it is orientated.
The answer is 30. This is the way I did it. I notionally numbered
the colours 1 to 6 and chose number one as a vertical face facing
me. This could always be the case. Then I asked myself 'where could
I place colour number 2?' It could be adjacent to the face coloured
1 or it could be opposite. Taking the adjacent case first, the cube
could be orientated by rotation so that the face coloured 2 was
on top leaving face 1 facing me. This fixes the orientation and
there are 4 faces left to colour so the number of colouring must
be 4! (factorial 4 = 24). Considering the case when the face coloured
2 is opposite, the cube is not fixed in orientation as it can rotate
without changing the assumptions so we can colour one of the other
faces with colour number 3 and rotate the cube so that this face
is uppermost. This will fix the orientation, leaving three faces
to colour. The permutations this time are 3! = 6. These 6 are independent
of the 24 we have already found so the answer is 6 + 24 = 30.
If F = the number of faces; E = the number of edges; V = the number
of vertices (corners), the full set of five regular Platonic solids
Tetrahedron (F=4; E=6; V=4)
Octahedron (F=8;E=12; V=6)
Cube (F=6; E=12; V=8)
Icosahedron (F=20; E=30; V=12)
Dodecahedron (F=12; E=30; V=20)
Note that in every case, V+F=E+2. This is known as Euler's theorem.
The number of edges that bound each face is P=2E/F.
My first question this month is very easy. How many ways can the
tetrahedron faces be coloured using four colours? Same rules as
My next question is about the octahedron which some people have
difficulty in visualising. Think of it like this: place four equilateral
triangles side by side so that they form a square pyramid. Now create
a second square pyramid in your mind and glue their bases together.
You now have an octahedron with eight triangular faces.
The octahedron is the dual of the cube which means that the vertices
are replaced by faces and vice versa. So you could, instead, colour
the corners of the cube rather than the faces of the octahedron.
(They have to be the same answer!)
So, in how many ways can you colour the faces of an octahedron
with eight different colours, making sure that no two colourings
are the same however orientated?
That's two questions in one month! New solvers need only answer
the first easy one about the tetrahedron: regular solvers must do
I hope that the puzzle this month will get you all thinking about
platonic solids and permutations but if not, you may have found
the puzzle page informative if nothing else.
Please send your answers to me, David Broughton, to arrive by 6th