Question

The number of straight lines joining 8 points on a circle is______.
A. $8$
B. $16$
C. $24$
D. $28$

Answer 1

Hint: Any two points can be joined to create a straight line. As a result, the number of straight lines that can be generated is a combination of choosing two points from 8 points on the circle.

Complete step by step solution:
We are given that all 8 points are on the circle. Hence these points are not collinear.
A straight line can be drawn by joining any 2 points.
Hence, the number of straight lines formed will be equal to the number of ways 2 points can be chosen from the 8 points on the circle.
We know that the number of ways of choosing r objects from n objects is given by $n{C_r}$.
Therefore, the number of ways we can choose 2 points from 8 points is given by $8{C_2}$
So, the number of straight lines that can be formed by joining 8 points is $8{C_2}$
Since $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $n! = n \times (n - 1) \times (n - 2) \times ...... \times 3 \times 2 \times 1$ we have,
$8{C_2} = \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}}$
$ \Rightarrow 8{C_2} = \dfrac{{8!}}{{2!6!}}$
$ \Rightarrow 8{C_2} = \dfrac{{8 \times 7}}{2}$
$ \Rightarrow 8{C_2} = 28$

Option ‘D’ is correct

Note: In order to solve the given question, one must know to form and calculate combinations.
The number of ways of selecting r objects from n objects is given by $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $n! = n \times (n - 1) \times (n - 2) \times ...... \times 3 \times 2 \times 1$.
One must also note that a straight line is formed by joining two points.