# Burnside's Lemma Triangle

I'm doing research into triangles and I was looking for the amount of possible triangles if 4 links are available between each point, without triangles which look the same if rotated. The 4 links possible are: no link, directed link outwards, directed link inwards, directed link outwards and inwards

For this I used Burnside's Lemma:

$\text{Number of orbits}=\frac{1}{|G|}\sum\limits_{g\in G} |X^g|$ = $\frac{1}{3}(|X^0|+|X^{120}|+|X^{240}|)$=$\frac{1}{3}(4^3+4+4)=24$

But if I draw all the triangles myself and the catagorise them I count 16 different possible triangles:

Link:16 triangles (I can't post images yet)

Did I use Burnside's Lemma wrong?

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You're using it right. There should be $24$ triangles that are distinct up to rotations. You are missing some that are related by reflection to figures that you have included, e.g., figures 9 and 10. If you want to exclude those too, then the group $G$ in Burnside's lemma needs to be $S_3$ instead of $\mathbb{Z}_3$. – mjqxxxx Jan 4 '13 at 16:33