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?

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

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

To solve this problem, we need to find the total number of different circles that can be drawn using any three of the given five points.

To begin, let's visualize the scenario. We have five points in the same plane, and we know that no three points are on the same line. This means that any three of these points will always form a unique triangle.

Now, to draw a circle through any three points, we need the points to be non-collinear, meaning they should not lie on a straight line. Since our points satisfy this condition, we can form different triangles from the given five points and draw circles passing through them.

To calculate the total number of different circles, we need to determine how many different triangles can be formed using the five points.

The formula to calculate the number of triangles formed from n points is given by:
C(n, 3) = n! / (3! * (n-3)!)

In this case, we have five points, so n=5. Substituting the values into the formula, we get:
C(5, 3) = 5! / (3! * (5-3)!)

Simplifying further:
C(5, 3) = 5! / (3! * 2!)
C(5, 3) = (5 * 4 * 3!) / (3! * 2)
C(5, 3) = (5 * 4) / 2
C(5, 3) = 10

Therefore, we can draw a total of 10 different circles, each passing through three of the given five points.