Permutations Practice Test 2

Consider the vertices of octagon as ${\mathit{A}}_1,{\mathit{A}}_2,...,{\mathit{A}}_8$
Triangles having only one side of the triangle as side of the octagon.
Consider this side as ${\mathit{A}}_1{\mathit{A}}_2$. Then third vertex may be any one from ${\mathit{A}}_4,{\mathit{A}}_5,{\mathit{A}}_6,{\mathit{A}}_7$ (adjacent vertices ${\mathit{A}}_3$ and ${\mathit{A}}_8$ are to be excluded). So number of triangles with one side ${\mathit{A}}_1{\mathit{A}}_2=4$.
∴ 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 $=8,\left[{\mathit{A}}_1{\mathit{A}}_2{\mathit{A}}_3,{\mathit{A}}_2{\mathit{A}}_3{\mathit{A}}_4,...,{\mathit{A}}_3{\mathit{A}}_1{\mathit{A}}_2\right]$
Also the number of all possible triangles ${=}^8{\mathit{C}}_3=56$
Hence, required number = 56 – 32 – 8 = 16.