Originally Posted by

**Archie Meade** Soroban was referring to the use of non-collinear points (no 3 points on a straight line).

In your image, you have 5 points.

If no 3 were collinear, you'd have 5C3 = 10 triangles.

However, in your diagram, 3 points are collinear along both diagonals,

so you lose 2 of those 10.

Unfortunately, where the diagonals cross has to be counted as a 5th point,

since you are using it in the formation of half of those 8 triangles.

(The formula counts triangles connected by points,

not including the formation of new points where lines cross).