Question

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. $116$
• B. $117$
• C. $120$
• D. $115$

Answer 1

The correct option is B $117$

Selection of any $3$ points from the $10$ points will define a circle. This can be done in ${}^{10}{C}_3$ ways.
And it is given that $4$ points are concyclic.
Hence, the required permutations will be
${}^{10}{C}_3-{}^4{C}_3+1$
$=120-4+1$
$=117$