Step 1: Understand the problem setup.
There are 10 points in a plane, 6 of which are collinear (and no other set of 3 or more points from the remaining 4 are collinear). We need to count the number of triangles that can be formed.
Step 2: Count all possible triangles ignoring the collinearity condition.
Any 3 points from 10 form a triangle, unless they are collinear. Total selections $= \binom{10}{3} = \dfrac{10 \times 9 \times 8}{6} = 120$.
Step 3: Subtract the degenerate cases.
The 6 collinear points give $\binom{6}{3}$ sets of 3 that are collinear (they do NOT form a triangle). $\binom{6}{3} = \dfrac{6 \times 5 \times 4}{6} = 20$.
Step 4: Apply the formula.
$N = \binom{10}{3} - \binom{6}{3} = 120 - 20 = 100$.
Step 5: Check if there are other collinear triples.
The remaining 4 points (the non-collinear ones) are in general position, so no 3 of them (or any mix with the collinear group that isn't already accounted for) are collinear. Thus no further subtractions are needed.
Step 6: State the answer.
The total number of triangles that can be formed is $120 - 20 = 100$. \[ \boxed{100} \]