Five points lie in the same plane with no three points on the same line. A circle can be drawn through any three of the five points. What is the total number of different circle that can be drawn such that each circle passes through three of the five points?

Answer 1

how many combinations of three from 5
n!/[ r!(n-r)! ]
5!/[2!(3!)] = 10

or do one by 1 the hard way

A B - C, D, E --> 3
A C - D, E -->2
A D - E -->1

B C - D, E -->2
B D - E -->1

C D - A -->1

3+2+1+2+1+1 = 10

Answer 2

Since no choice of 3 points forms a straight line, you are simply finding the number of ways to choose 3 of the 5 points, which is
C(5,3) or 5!/(3!2!) = 10