Question

Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the maximum number of circles that can be drawn so that each contains at least three of the given points is
A. 216
B. 156
C. 172
D. None

Answer 1

In the above given question, we are given 11 points, out of which 5 points lie on the same one circle. Out of the rest of 6 points, no 4 points lie on the same circle. That means out of the 6 other points, only a maximum of 3 points can lie on the same circle. We have to determine the maximum number of circles that can be drawn using the given 11 points such that each circle contains at least three of the given 11 points.

Complete step by step answer:
In order to approach the solution, we must consider the property of a circle that if three different points lie on a circle, then that circle is unique. That means those three points can not lie on another circle.Therefore, to find the maximum number of circles that can be drawn using the given 11 points such that each circle contains at least three of given 11 points, we have to choose at least three points at random out of these 11 points.

We must consider the given information that 5 out of 11 points lie on a same circle, while out of the other 6 points, no 4 points lie on the same circle.Hence, let us separate these points in two groups i.e. the first 5 points and the 6 other points, while choosing three points out of them.Now, we only have to find the number of ways we can choose 3 points out of them.We can choose those 3 points out of the 6 points in $^6{C_3}$ ways. Or, we can choose 1 point from the 6 points and another 2 points from the 5 points in $^6{C_1}^5{C_2}$ ways. Or, the vice-versa of the above way in $^6{C_2}^5{C_1}$ ways.But if we choose all 3 points out of the 5 points, which lie on the same circle, then there is only $1$ way because that circle is unique.

Now, we have to add all the possible number of ways we can choose the required circles.
That is given by,
${ \Rightarrow ^6}{C_3}{ + ^6}{C_1}^5{C_2}{ + ^6}{C_2}^5{C_1} + 1$
That gives us the expression,
$\Rightarrow \left( {\dfrac{6}{3} \cdot \dfrac{5}{2} \cdot \dfrac{4}{1}} \right) + \left( {\dfrac{6}{1} \times \dfrac{5}{2} \cdot \dfrac{4}{1}} \right) + \left( {\dfrac{6}{2} \cdot \dfrac{5}{1} \times \dfrac{5}{1}} \right) + 1$
Solving this sum, we get
$\Rightarrow 20 + 60 + 75 + 1$
That is,
$\Rightarrow 156$
That is the required number of ways. Therefore, the maximum number of circles that can be drawn so that each contains at least three of the given points is $156$.

So the correct option is B.

Note: A combination is the selection of all or part of a set of objects, without regard to the order in which objects are selected and is denoted by the expression $^n{C_r}$. Here $^n{C_r}$ is the number of combinations, that shows that if we have to select only $r$ random objects from a set of total $n$ objects, then we get a total number of combinations $^n{C_r}$ which is given by the formula,
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$