Answer these questions based on the following informations
Two parallel lines each have a number of distinct points marked on them. On one line there are two points P and Q. On the other line there are eight points
Find the maximum number of different triangles which could be formed having three of the ten points as their vertices

Options

A. 3
B. 2
C. 64
D. 6

Answer

To solve the problem of finding the maximum number of different triangles that can be formed using the points on two parallel lines, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Identify the Points**:
- On the first line, we have 2 distinct points: P and Q.
- On the second line, we have 8 distinct points: A1, A2, A3, A4, A5, A6, A7, A8.

2. **Understand Triangle Formation**:
- A triangle can be formed by selecting 3 points. However, since the points are on two parallel lines, we need to ensure that not all three points are on the same line.
- Therefore, we can form triangles in two distinct cases:
- **Case 1**: 1 point from the first line and 2 points from the second line.
- **Case 2**: 2 points from the first line and 1 point from the second line.

3. **Calculate for Case 1**:
- We need to select 1 point from the first line (which has 2 points) and 2 points from the second line (which has 8 points).
- The number of ways to select 1 point from 2 is given by the combination formula \( \binom{n}{r} \):
\[ \text{Ways to choose 1 from 2} = \binom{2}{1} = 2 \]
- The number of ways to select 2 points from 8 is:
\[ \text{Ways to choose 2 from 8} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
- Therefore, the total number of triangles for Case 1 is:
\[ \text{Total for Case 1} = 2 \times 28 = 56 \]

4. **Calculate for Case 2**:
- We need to select 2 points from the first line (which has 2 points) and 1 point from the second line (which has 8 points).
- The number of ways to select 2 points from 2 is:
\[ \text{Ways to choose 2 from 2} = \binom{2}{2} = 1 \]
- The number of ways to select 1 point from 8 is:
\[ \text{Ways to choose 1 from 8} = \binom{8}{1} = 8 \]
- Therefore, the total number of triangles for Case 2 is:
\[ \text{Total for Case 2} = 1 \times 8 = 8 \]

5. **Combine Both Cases**:
- Now, we add the number of triangles from both cases to find the total number of different triangles:
\[ \text{Total triangles} = \text{Total for Case 1} + \text{Total for Case 2} = 56 + 8 = 64 \]

6. **Final Answer**:
- The maximum number of different triangles that can be formed is **64**.