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 task is to determine the number of triangles formable from 12 points on a plane, given that 5 of these points are collinear. The solution proceeds as follows:

1. Initially, calculate the total possible combinations of selecting 3 points from the 12 available points. This is computed using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items and \( r \) is the number of items to choose:
   • \( \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \)
2. Since 5 points lie on the same line (collinear), any selection of 3 points from this group will not form a triangle. These combinations must be excluded. The number of such combinations is calculated as:
   • \( \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \)
3. To find the actual number of triangles, subtract the collinear combinations from the total combinations:
   • Total triangles = Total combinations of 3 points - Collinear combinations
   • Total triangles = \( 220 - 10 = 210 \)
4. Consequently, the total number of triangles that can be formed under the specified conditions is \(210\).

Therefore, the definitive answer is \(210\), corresponding to option \(\displaystyle (a)\).