Question

How many triangles can I draw with n non-colinear dots?

be 'n' a Natural number

Answer 1

Total number of triangles that could be drawn is `(n(n-1)(n-2))/6`.

Explanation:

As one can draw a triangle with any three non-collinear dots, we have to choose any three points out of total `n` non-collinear points to make a triangle.

Observe that order of selection does not matter. As we can choose the `3` points out of `n` points in `C_3^n` ways, the total number of triangles that could be drawn is

`C_3^n=(n(n-1)(n-2))/(1*2*3)=(n(n-1)(n-2))/6`

Note that `C_r^n=(n(n-1)(n-2)......(n-r+1))/(1*2*3....r)`.