Question

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:

• A. \(210\)
• B. \(200\)
• C. \(230\)
• D. \(220\)

Hint:

When calculating the number of triangles, subtract the cases where all three points are collinear from the total number of combinations.

Answer 1

The Correct Option is A

The total number of ways to choose 3 points out of 12 is given by the combination formula: \[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \] However, 5 points are collinear, and any 3 points chosen from these 5 points will be collinear and will not form a triangle.
The number of ways to choose 3 points out of 5 collinear points is: \[ \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \] So, the total number of triangles that can be formed is: \[ \binom{12}{3} - \binom{5}{3} = 220 - 10 = 210 \]