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

<p>To count the number of distinct triangles:</p> <p><p><strong>Start with all possible triples of points.</strong> </p> <p>From 12 points, the number of ways to choose any 3 is </p> <p>$ \binom{12}{3}= \frac{12\cdot11\cdot10}{3\cdot2\cdot1}=220. $</p></p> <p><p><strong>Subtract the “invalid” triples that are collinear.</strong> </p> <p>The only collinear sets of three points arise from the single line containing the 5 collinear points. </p> <p>Number of ways to pick 3 points from those 5 is </p> <p>$ \binom{5}{3}=10. $</p></p> <p><p><strong>Valid triangles = total triples − collinear triples.</strong> </p> <p>$ 220-10 = 210. $</p></p> <p>Hence, the total number of triangles that can be formed is <strong>210</strong>.</p> <p><strong>Correct option: B</strong></p>