Question

There are ${\text{n}}$points on a circle. The number of straight lines formed by joining them is equal to
A. ${}^{\text{n}}{{\text{C}}_{\text{2}}}$
B. ${}^{\text{n}}{{\text{P}}_{\text{2}}}$
C. ${}^{\text{n}}{{\text{C}}_{\text{2}}}{\text{ - 1}}$
D. None of these

Answer 1

Hint: A straight line can be formed by joining any two points. So, the number of straight lines that can be formed is the combination of selecting 2 points from ${\text{n}}$ points on the circle.

Complete step by step answer:

As the ${\text{n}}$ points are on a circle, a straight line can be drawn by connecting any 2 points. So, the number of straight lines formed will be equal to the number of ways 2 points can be selected from the ${\text{n}}$ points on the circle.
The number of ways of selecting r objects from n objects is given by ${}^n{C_r}$.
Therefore, the number of ways we can select 2 points from ${\text{n}}$ points is given by ${}^{\text{n}}{{\text{C}}_{\text{2}}}$.
So, the number of straight lines that can be formed by joining ${\text{n}}$ points is ${}^{\text{n}}{{\text{C}}_{\text{2}}}$
Therefore, the correct answer is option A.

Note: It is a basic concept in geometry that the line passing through 2 points will be unique. As the ${\text{n}}$points are on a circle, we cannot draw a line connecting more than 2 points. So, every line in that can be formed by connecting 2 points on a will be unique. So, we need to select 2 points out of n points. For that we use the concept of combinations. The number of ways of selecting r objects from n objects is given by ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. We are not using permutations as the order of the points selected is not important. When we join two points P and Q, we get the lines PQ and QP. But QP and PQ represent the same line. If we take permutations, PQ and QP will be considered as two lines. so we cannot take permutations.