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Given 11 points, of which 5 lie on one circle, other than these 5, no

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Bunuel
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Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink] New post   22 Feb 2021, 09:45
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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 at least three of the given points is:

(A) 216
(B) 156
(C) 142
(D) 140
(E) 135



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sujoykrdatta
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Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink] New post   22 Feb 2021, 13:14
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Bunuel wrote:
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) 142
(D) 140
(E) 135



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A circle (unique) can always be drawn through 3 points (non-collinear)
Ideally, using the 11 points, we should have got 11c3 = 165 circles.
But, the 5 points which are already in a single circle, would have been counted as 5c3 = 10 circles.
We need to remove these 10 and instead include only 1
Thus, total = 165 - 10 + 1 = 156

Answer B

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Re: Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink] New post   13 Mar 2021, 16:42
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Bunuel wrote:
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) 142
(D) 140
(E) 135




Solution:

Since we want the maximum number of circles that can be drawn, we assume that any 3 points we choose will be capable of creating a circle.

Thus, we have 11C3 = 11! / (3! x 8!) = 165 possible circles, each with 3 points from the 11.

However, one circle has 5 points. Those 5 points could have been used to create 5C3 = 10 circles, but instead, they were used to make only 1 circle. Thus, we have to subtract 9 from the 165 possible 3-point circles, yielding 165 - 9 = 156.

Answer: B
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Re: Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink] New post   26 Apr 2021, 14:28
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:

Nos. of circles without any restrictions = 11C3 = 165

Now, 5 lies on a circle = 5C3 = 10 and it makes 1 circle only.

So, required nos. of circle = 165-10+1= 156

it is C. :)
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Re: Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink] New post   27 Apr 2021, 04:24
The direct method is not as efficient but here it goes;

1+ 6C3+ (5C2x6C1)+(5C1x6C2)=156

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Re: Given 11 points, of which 5 lie on one circle, other than these 5, no [#permalink]
27 Apr 2021, 04:24
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