Question

The numbers of lines drawn through 6 points lying on a circle is.
(A) $ 12 $
(B) $ 15 $
(C) $ 24 $
(D) $ 30 $

Answer 1

Hint: A line is drawn using two points. Use this concept in combination to solve the above question. Arrangement is represented by permutation and selection is represented by combination. You need to be very careful while deciding whether you should use permutation or combination.

Complete step-by-step answer:
We know that a line can be drawn by joining any two points.
Therefore, two draw a line from the 6 given points, we need to select any 2 points.
Selection of $ r $ different objects from $ n $ different objects is given by $ {}^n{C_r} $
Where,
 $ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $
Where, $ n $ is the total number of distinct available objects.
 $ r $ is the total number of distinct objects that we need to select
 $ \Rightarrow n! = n(n - 1)(n - 2).....3 \times 2 \times 1 $
Therefore, selection of 2 points out of 6 given points is given by
 $ \Rightarrow {}^6{C_2} = \dfrac{{6!}}{{2!(6 - 2)!}} $
 $ = \dfrac{{6 \times 5 \times 4!}}{{2 \times 1 \times 4!}} $ $ \left( {\because n! = n(n - 1)!} \right) $
Simplifying it, we get
 $ \Rightarrow {}^6{C_2} = 15 $
Therefore, we can draw 15 lines from the 6 points lying on a circle.
Therefore, from the above explanation, the correct answer is, option (B) 15

So, the correct answer is “Option B”.

Note: Selection and arrangement are two different things. In arrangement, sequence matters. But in selection, sequence does not matter.
For example, we can arrange A and B as AB or BA. But if we want to select A and B, then it does not matter if I select A first or B first. That means, I have two ways of arranging A and B but I have only one way of selecting A and B.