There are a few conflicting answers here. The correct count is 16. One approach (in a comment by @juantheron) is to consider all triples of vertices and subtract those that share exactly one side with the octagon and those that share exactly two sides with the octagon: $$\binom{8}{3} - 8\cdot 4 - 8 = 56 - 32 - 8 = 16.$$ A similar approach is to use the inclusion-exclusion principle. Start with all triples of vertices, subtract (an overcount of) those that share at least one side, and add back in those that share two sides: $$\binom{8}{3} - 8\cdot 6 + 8 = 56 - 48 + 8 = 16.$$ Here is the explicit list of 16: ACE, ACF, ACG, ADF, ADG, AEG, BDF, BDG, BDH, BEG, BEH, BFH, CEG, CEH, CFH, DFH Anyone who claims a count higher than 16, please show me one that is not on this list. Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now |