Question

The number of triangles whose vertices are at the vertices of an octagon but none of whose side happen to come from the sides of the octagon is

• A. 24
• B. 52
• C. 48
• D. 16

Answer 1

Answer: (D)

16

Answer 2

Number of all possible triangles = Number of selections of 3 points from 8 vertices $=^8C_3=56$ Number of triangle with one side common with octagon $= 8 ? 4 = 32$ (Consider side $A_1A_2$. Since two points $A_3, A_8$ are adjacent, 3rd point should be chosen from remaining 4 points.) Number of triangles having two sides common with octagon : All such triangles have three consecutive vertices, viz., $A_1A_2A_3, A_2A_3A_4, ..... A_8A_1A_2.$ Number of such triangles $= 8$ $\therefore$ Number of triangles with no side common $= 56 - 32 - 8 = 16.$