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 atleast three of the given points is

216

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156

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172

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None

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Solution

The correct option is B $156$
Let the no. of 5 points on the circle be represented as $p_{1}, p_{2}, p_{3}, p_{4}$ and $p_{5}$ .
$=^{6} C_{3} + (^{6} C_{1} \cdot^{5} C_{2}) + (^{6} C_{2} \cdot^{5} C_{1}$ )+1
$= \frac{6 \times 5 \times 4}{6} + (6 \times 10) + (15 \times 5) + 1$
$= 156$

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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 atleast three of the given points is

The correct option is B 156Let the no. of 5 points on the circle be represented as p1,p2,p3,p4 and p5.=6C3+(6C1⋅5C2)+(6C2⋅5C1)+1=6×5×46+(6×10)+(15×5)+1=156

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 atleast three of the given points is

The correct option is B $156$
Let the no. of 5 points on the circle be represented as $p_{1}, p_{2}, p_{3}, p_{4}$ and $p_{5}$ .
$=^{6} C_{3} + (^{6} C_{1} \cdot^{5} C_{2}) + (^{6} C_{2} \cdot^{5} C_{1}$ )+1
$= \frac{6 \times 5 \times 4}{6} + (6 \times 10) + (15 \times 5) + 1$
$= 156$