There are n points in a plane, No three being collinear except m of them which are collinear. The number of triangles that can be drawn with their vertices at three of the given points is

Options

A. `""^(n-m)C_3`
B. `""^(n)C_(3) -""^(m)C_3`
C. `""^(n)C_(3)-m`
D. None of these

Answer

To solve the problem of finding the number of triangles that can be formed with vertices at three of the given points, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Total Points**: We have a total of \( n \) points in the plane.

2. **Identify Collinear Points**: Among these points, \( m \) points are collinear. This means that these \( m \) points cannot form a triangle.

3. **Calculate Total Combinations of Points**: The total number of ways to choose any 3 points from \( n \) points is given by the combination formula:
\[ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]

4. **Calculate Combinations of Collinear Points**: The number of ways to choose 3 points from the \( m \) collinear points (which cannot form a triangle) is:
\[ \binom{m}{3} = \frac{m(m-1)(m-2)}{6} \]

5. **Calculate Valid Triangles**: The number of triangles that can be formed is the total combinations minus the combinations of collinear points:
\[ \text{Number of triangles} = \binom{n}{3} - \binom{m}{3} \]

### Final Formula:
Thus, the final formula for the number of triangles that can be formed is:
\[ \text{Number of triangles} = \frac{n(n-1)(n-2)}{6} - \frac{m(m-1)(m-2)}{6} \]