Problem 623

Since there are four triangular sections in each flag, then at most four colours can be used in a single flag.

Since no two adjacent triangles are the same colour, then at least two colours must be used. (For example, the sections Top and Left must be different colours.)

Therefore, the number of colours used is 2, 3 or 4.

We count the number of possible flags in each case.

Case 1: 2 colours

We call the colours A and B.

Assign the colour A to Top.

Since Left and Right cannot be coloured A and there is only one other colour, then Left and Right are both coloured B.

Bottom cannot be coloured B (since it shares an edge with Left and Right) so must be coloured A.

This gives us:

[[IMAGE0]]

This configuration does not violate the given rule.

There are 5 possible colours for A (red, white, blue, green, purple).

For each of these 5 choices, there are 4 possible colours for B (any of the remaining 4 colours).

Therefore, there are $5(4) = 20$ possible flags in this case.

Case 2: 4 colours

We call the colours A, B, C, and D.

Since there are 4 sections and 4 colours used, then each section is a different colour.

We label them as shown:

[[IMAGE1]]

This configuration does not violate the given rule.

There are 5 possible colours for A. For each of these 5 choices, there are 4 possible colours for B. For each of these combinations, there are 3 possible colours for C and 2 possible colours for D.

Therefore, there are $5(4)(3)(2)= 120$ possible flags in this case.

Case 3: 3 colours

We call the colours A, B and C.

Assign the colour A to Top.

Since Left cannot be coloured A, we assign it the colour B.

Section Right cannot be coloured A, so could be B or C.

If Right is coloured B, then in order to use all three colours, Bottom must be coloured C.

If Right is coloured C, then Bottom (which shares an edge with each of Left and Right) must be coloured A.

This gives two possible configurations:

[[IMAGE2]] [[IMAGE3]]

Neither configuration violates the given rule.

In each configuration, there are 5 possible colours for A. For each of these 5 choices, there are 4 possible colours for B. For each of these combinations, there are 3 possible colours for C.

Since there are two such configurations, then there are $2(5)(4)(3) = 120$ possible flags in this case.

In total, there are $20+120+120=260$ possible flags.