What is the number of triangles that can be formed using vertices are the vertices of an octagon but have only one side common with that of octagon?

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.

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