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22 Feb 2021, 09:45
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B
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Difficulty:
75%
(hard)
Question Stats:
46% (01:57) correct
54%
(02:18)
wrong
based on 92
sessions
History
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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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(A) 216
(B) 156
(C) 142
(D) 140
(E) 135
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
22 Feb 2021, 13:14
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Bunuel
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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General Discussion
13 Mar 2021, 16:42
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Bunuel
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
