There are 20 points all in one straight line and 15 others in a parallel line. How many possible triangles having these points as vertices can be formed?
I can form a triangle by choosing any two points on the one line, then one of the points from the other line.
so the number of ways
= C(20,2)x15 + C(15,2)x20
= 2850 + 2100
= 4950
To solve this problem, we need to determine the number of ways we can choose three points from the given set of points to form a triangle.
First, let's consider choosing three points from the 20 points on the straight line. Since the points are collinear, we can't form a triangle with these points alone, as they won't enclose any area.
Next, let's consider choosing two points from the 20 points on the straight line and one point from the 15 points on the parallel line. We have 20 possibilities for the first point, 19 possibilities for the second point (as we cannot choose the same point twice), and 15 possibilities for the point on the parallel line. Therefore, there are 20 * 19 * 15 = 5,700 possible triangles formed this way.
Finally, let's consider choosing three points from the 15 points on the parallel line. Since these points are also collinear with the other set, we can't form a triangle with just these points. Therefore, there are no triangles formed this way.
Adding up the two cases, we find that there are a total of 5,700 possible triangles that can be formed using these points.