Greg uses combination formula to solve this problem. He says (7C2 * 5) + (5C2 * 7). But 7C2 includes any 2 collinear points from the bottom row. It can even be the 1st point in the bottom row and the last point in the bottom row. That means we are actually using all the points in the bottom row to form a triangle when the question clearly says to use only 3 points. So, is it really appropriate to use the combinations formula for this question?
My approach would be to include only consecutive two points from bottom and one point from top row (6 X 5). Similarly, consecutive two points from top and one point from bottom row
(4 X 7).Total = 30+28=58. What am I missing?
Well nowhere does it say that you’re only allowed to pick three points. I think you’re reading “connecting three of the points” independently without looking at the “as vertices of the triangle” part, which leads to confusion.
Essentially, you want to pick three vertices, and it doesn’t matter if you cross the other points in the process of doing that.
The way you parsed the question (but you seem to have misinterpreted), your answer is not wrong.