Permutations Practice Test 2

Consider the vertices of octagon as A1,A2,...,A8
Triangles having only one side of the triangle as side of the octagon.
Consider this side as A1A2. Then third vertex may be any one from A4,A5,A6,A7 (adjacent vertices A3 and A8 are to be excluded). So number of triangles with one side A1A2=4.
∴ Number of triangles with one side as side of octagon = 8×4 = 32.
Now the number of triangles having 2 sides of octagon as their sides =8,[A1A2A3,A2A3A4,...,A3A1A2]
Also the number of all possible triangles =8C3=56
Hence, required number = 56 – 32 – 8 = 16.