Given $n$ different points in a plane. 
Given $n$ different points in a plane, $8$ of them are on one straight
line. The other points are in general everywhere else, so there are no
$3$ points on same one straight line. How many different triangles can
you create from these n points?

What I think is: since no three points in the plane are collinear, a triangle can be formed by selecting any three of the $n$ points. The three points can be selected in $^nC_3$ ways, similarly, the number of triangles formed by $8$ collinear points when no three points are collinear in the plane is $^8C_3$. However, the triangles formed by these points are not allowed, so the total number of triangles formed would be ${^nC_3}- {^8C_3}$
Is that correct?

As has been stated in the comments, your solution is correct; $\binom n3-\binom 83$ different non-degenerate triangles can be formed from this points.
