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

Updated On: Jul 7, 2022
  • 24
  • 52
  • 48
  • 16
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The Correct Option is D

Solution and Explanation

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 $= 56 - 32 - 8 = 16.$
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Notes on Permutations

Concepts Used:

Permutations

A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:

\(^nP_r = \frac{n!}{(n-r)!}\)

 nPr = permutation

 n = total number of objects

 r = number of objects selected

Types of Permutation

  • Permutation of n different things where repeating is not allowed
  • Permutation of n different things where repeating is allowed
  • Permutation of similar kinds or duplicate objects