Question

Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beam with the top of its non-adjacent pillars, then the total number of beams is:
A) 210
B) 190
C) 170
D) 180

Answer 1

Hint: In this to find the total number of beams. We will consider the circular stadium to be a n-sides polygon and the number of vertices of the polygon. Since there are 20 pillars therefore we will have a 20-sided polygon. Since each beam is attached to two non-adjacent pillars. Therefore we have to find the number of diagonals of a 20-sided polygon.
n!= n x (n-1) x (n-2) x…x 2 x 1. i.e. factorial of n is the product of all n natural numbers starting from 1 to n.

Complete step-by-step answer:
Given that, there are 20 pillars of the same height erected along the boundary in the circular stadium.
Since the top of two non-adjacent pillars is attached to a beam
Therefore, if we consider a circular stadium with 20 pillars (vertices of polygon) of same height as a 20-sides polygon and the beam attached to pillars as a diagonal of polygon. Then total number of beams = total number of diagonals to 20 sides polygon
Number of selection of section of two vertices of polygon= ${}^{20}{{\text{C}}_{2}}$(includes adjacent vertices also)
Number of selection of section of two non-adjacent vertices of polygon = ${}^{20}{{\text{C}}_{2}}-20$
Total number of diagonals to 20 sided polygon = ${}^{20}{{\text{C}}_{2}}-20$
Total number of diagonals to 20 side’s polygon = $\dfrac{20!}{2!\left( 20-2 \right)!}-20=\dfrac{20!}{2!\cdot 18!}-20$
Total number of diagonals to 20 side’s polygon $=\dfrac{20\times 19\times 18!}{2\times 1\times 18!}-20=\dfrac{20\times 19}{2}-20$
Total number of diagonals to 20 side’s polygon = 10 X 19 – 20 = 190 – 20 = 170
Therefore the total number of beams is 170.

So, the correct answer is “Option C”.

Note: In this problem, one remember that we are selecting non adjacent vertices (pillars). We used the combination for the selection of the vertices. Always remember that ${}^{n}{{\text{C}}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, where $n!=n(n-1)(n-2)......3.2.1$ . Try not to make any calculation mistakes.