How many distinguishable colourings of a regular triangle are there if each side can be painted one of two colours.
X is the set of all colourings of a regular triangle where each side can be coloured one of two colours. (S3,*) is the symmetry group of a regular triangle. Substitution tells you the answer is 24/6 which is 4.
This is a combinatorics problem. The ordered set X of all colourings of a triangle (regular or irregular) where each side can be coloured one of two colours is represented by the following sets, using 1 and 2 for the two colours - {1,1,1}, {1,1,2}, {1,2,1}, {2,1,1}, {1,2,2}, {2,1,2}, {2,2,1}, {2,2,2}. Indeed, being such a colouring expert, you obviously know this is the same as the number of ways of colouring any 3 things with 2 colours, whether they are triangles, have symmetry or not.
You'll note that there are only 8 possible combinations, rather then the 24 you've somehow arrived at above.
The number of symmetries isn't applicable here - "distinguishable" means that a triangle with 1 red and 2 green sides is the same as another, whichever side we choose as the red one. We reduce the ordered set to an unordered one, giving {1,1,1}, {1,1,2}, {1,2,2} and {2,2,2}.
So, well done on getting the right answer (it's not that hard), but your reasoning is rubbish. You said you used Burnside's Colouring Formula to get your answer, but your working doesn't show that at all. Given the simplicity of the question you posed, it would be ridiculous overkill to even try to apply a formula aimed at much more complex scenarios, when simple maths will cover it.
You did a dissertation as part of your bachelors degree, you say? That's very unusual. Where did you study?
I'm sure I speak for all those interested in graph and network theory to read "Groups and an Application to Colouring Problems". Why don't you post it for us, and remove all doubt about your expertise?