There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through these 10 points is
A
116
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B
117
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C
120
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D
115
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Solution
The correct option is B117 Selection of any 3 points from the 10 points will define a circle. This can be done in 10C3 ways. And it is given that 4 points are concyclic. Hence, the required permutations will be 10C3−4C3+1 =120−4+1 =117