Htykuut

I'm referring to problem A6 of the 2017 Putnam competition -- the question is "How many ways exist to colour the labelled edges of an icosahedron such that every face has two edges of the same colour and one edge of another colour, where the colours are either red, white or blue?".

My solution is as follows: consider the planar representation of the icosahedron:

Note that:

• There are 18 ways to colour the edges of a triangle such that it has two edges of the same colour and one edge of another colour (3 choices for the colour that appears twice, 2 choices for the colour that appears once, and 3 choices for the arrangement).




• Given the colouring on any one edge of a triangle, there are 6 ways to colour the remaining edges in a way that satisfies the condition (WLOG suppose the given edge is white -- then either one other edge is white (and the other edge is red or blue), which has 4 possibilities, or both the other edges have the same colour, red or blue, which is possible in 2 ways.




• Given the colouring on two edges of a triangle, there are 2 ways to colour the remaining edge in a way that satisfies the condition. WLOG, the given edges are coloured either "R R" or "R B". If it's "R R", the 2 ways to choose the other edge are "W" and "B" -- if it's "R B", the 2 ways to choose the other edge are "R" and "B".

So there are 18 ways to choose the colouring on the central triangle (the base case) of the planar representation, and you can write each the number of ways to colour each successive "containing triangle" as $6^32^3$ multiplied by the smaller triangle it contains, so the number of ways to colour the entire icosahedron should be:

$$18(6^32^3)^3=2^{19}3^{11}$$

Unfortunately, the official solution (p. 5) presents an answer of $2^{20}3^{10}$ -- I'm off by a factor of $2/3$! What's going on? What did I do wrong?

You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. The rest seems o.k. to me.