On a triangle ABC, on the side AB, 5 points are marked, 6 points are marked on the side BC and 3 points are marked on the side AC (none of the points being the vertex of the triangle). How many triangles can be made by using these points?

Options

A. 90
B. 333
C. 328
D. None of these

Answer

To solve the problem of how many triangles can be formed using the marked points on the sides of triangle ABC, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Points**:
- On side AB, there are 5 points.
- On side BC, there are 6 points.
- On side AC, there are 3 points.

2. **Calculate Total Points**:
- Total points = Points on AB + Points on BC + Points on AC
- Total points = 5 + 6 + 3 = 14 points.

3. **Choose 3 Points to Form a Triangle**:
- To form a triangle, we need to select 3 points from the total of 14 points.
- The number of ways to choose 3 points from 14 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of points and \( r \) is the number of points to choose.
- Thus, we need to calculate \( \binom{14}{3} \).

4. **Calculate \( \binom{14}{3} \)**:
- Using the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \):
\[ \binom{14}{3} = \frac{14!}{3!(14-3)!} = \frac{14!}{3! \cdot 11!} \]
- This simplifies to:
\[ \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = \frac{2184}{6} = 364 \]

5. **Subtract Collinear Points**:
- We need to subtract the cases where the selected points are collinear (which do not form a triangle).
- Collinear points can be chosen from each side:
- For side AB (5 points): \( \binom{5}{3} = 10 \)
- For side BC (6 points): \( \binom{6}{3} = 20 \)
- For side AC (3 points): \( \binom{3}{3} = 1 \)
- Total collinear combinations = \( 10 + 20 + 1 = 31 \).

6. **Calculate Valid Triangles**:
- The total number of valid triangles is:
\[ \text{Total triangles} = \text{Total combinations} - \text{Collinear combinations} = 364 - 31 = 333. \]

### Final Answer:
The total number of triangles that can be formed is **333**.