How many different arrangements of triangles that are either red or blue around a regular heptagon are possible? I have the following problem I have an yellow heptagon (regular $7$ sided polygon) Against every side there is a triangle. The triangle is either red or blue. How many different arrangements of triangles around the heptagon are possible? arrangements that are rotational variants don't count, arrangements that are mirror variants can count so going clockwise and having counted red-red-red-blue-red-blue-blue as arrangement then red-red-blue-red-blue-blue-red is the same arrangement and does not count (it is the same arrangement rotated a bit) red-red-blue-blue-red-blue-red is not the same arrangement (but is a mirror) Related and how many when we also discard the mirror arrangements?

Since $7$ is prime, there are no periods other than $1$ and $7$. There are $2$ arrangements with period $1$, so the remaining $2^7-2$ arrangements have period $7$. Thus there are

$$ 2+\frac{2^7-2}7=20 $$

rotationally inequivalent arrangements.

The orbits under the full symmetry group including reflections can be counted using Burnside's lemma. The identity leaves $2^7$ arrangements invariant, each of $6$ rotations leaves $2$ arrangements invariant and each of $7$ reflections leaves $2^4$ arrangements invariant, so there are

$$ \frac{2^7+6\cdot2+7\cdot2^4}{14}=18 $$

inequivalent arrangements.