There are 10 points in a plane, no three of which are collinear, except for 4 points which are collinear. How many distinct triangles can be formed?

Solution

1. The total number of ways to choose 3 points from 10 is given by the combination formula $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{10}{3}\right)& =\frac{10\times 9\times 8}{3\times 2\times 1}\\ & =120\end{array}$$.
2. Since 4 points are collinear, any selection of 3 points from these 4 will fail to form a triangle, which occurs in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ ways.
3. Subtracting these invalid cases from the total gives $120 - 4 = 116$ possible triangles.
4. Hence the answer is (A).