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

Solution

Maths sir · 3 steps

Choose any 3 of the 12 points, then throw away the triples that sit on one line.

1. All triples. Any 3 points that are not collinear form a triangle. Ignoring the collinear set for now, the number of ways to pick 3 points from 12 is C(12, 3) = (12 × 11 × 10)/6 = 220.
2. Remove bad triples. Any 3 points chosen from the 5 collinear points lie on one straight line, so they give no triangle. There are C(5, 3) = 10 such triples.

   C(12, 3) - C(5, 3) = 220 - 10 = 210

3. So. The number of triangles is 210, which is option (b).

The idea to keep. Count every selection first, then subtract the selections that fail. With m collinear points among n, triangles = C(n, 3) - C(m, 3).

Answer from the NTA final answer key. Solution by PrepWiser.