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26 Feb 2012, 00:05
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B
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Difficulty:
25%
(medium)
Question Stats:
75% (01:20) correct
25%
(01:44)
wrong
based on 562
sessions
History
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Not Attempted Yet
Assume that there are 12 points on a straight line as shown below:
--*-*-*----*-*-*-*-*--*-*-*-*
How many circles can be drawn on the line segment above if each circle is subject to the rule that the endpoints of one of its diameters must be any two of the points?
36
66
72
78
121
--*-*-*----*-*-*-*-*--*-*-*-*
How many circles can be drawn on the line segment above if each circle is subject to the rule that the endpoints of one of its diameters must be any two of the points?
36
66
72
78
121
26 Feb 2012, 00:15
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Chembeti
Any 2 different points out of given 12 colinear points can be the endpoints of the diameter, so there can be \(C^2_{12}=66\) circles drawn.
Answer: B.
23 Oct 2013, 19:10
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This is an interesting way to disguise a combination problem. As Bunuel noted, you can create a circle through any two points, so what you really have is a question of how many unique ways to select pairs from a set of 12 points, or:
12!/2!(12-2)! --> 12!/2!10! --> (12*11)/2 --> 6*11 = 66
12!/2!(12-2)! --> 12!/2!10! --> (12*11)/2 --> 6*11 = 66
