Two lines are parallel. On the first line, there are 2 points, P and Q. On the second line, there are 6 points, A, B, C, D, E and F. How to calculate the number of triangles that can be formed by drawing straight lines that pass through any 3 points?
To form a triangle you need two points from one line and the other from the next line.
If you take the second line (with points A-F) as base and you get two points (AB,BC,AC etc) from that line, you will get the following combination.
Ways that you can choose two points `= ^6C_2 = 15`
In the other line you have two points (P and Q). So when we connect the above two points with P and Q we have `^2C_1 = 2` ways to do that.
So ways to perform triangles when we use second line as base is `15xx2 = 30` ways.
These triangles will looks like this.
ABP, ACP, ABQ,ADQ etc
When you choose first line as base you will have 6 points to connect with on the other line. So the number of ways would be `^6C_1 = 6`
These triangles will look like PQA,PQB,PQC etc.
So the total triangles can be formed are (30+6) = 36.