There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is:

Step 1: Calculate total possible triangles without collinearity

First, let's consider the total number of triangles that could be formed if no three points were collinear. With $n$ points, the number of ways to choose 3 points is given by the combination formula $\binom{n}{3}$. Here, $n=12$, so we calculate $\binom{12}{3}$.

Step 2: Calculate triangles from collinear points

Next, we identify the number of points that are collinear. The problem states that 5 points are collinear. If we choose any three points from these 5 collinear points, they will not form a triangle, but rather a straight line segment. We calculate this using $\binom{k}{3}$, where $k=5$.

Step 3: Subtract invalid triangles

To find the actual number of triangles, we subtract the number of 'triangles' formed by the collinear points from the total number of possible triangles. This is because the collinear points cannot form a triangle.

Step 4: Calculate combinations

We calculate the values of the combinations: $\binom{12}{3}$ and $\binom{5}{3}$. $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$ $\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 5 \times 2 = 10$

Step 5: Final calculation

Finally, we subtract the number of invalid triangles from the total possible triangles to get the correct count.

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