There are 10 points on a straight line AB and 8 on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices?

Options

A. 720
B. 640
C. 816
D. None of these

Answer

To solve the problem of how many triangles can be formed using the given points on lines AB and AC, we will follow these steps:

### Step 1: Understand the Problem
We have two lines:
- Line AB has 10 points.
- Line AC has 8 points.
We need to form triangles using these points without including point A.

### Step 2: Identify Triangle Formation Conditions
A triangle can be formed by choosing any three points, but since all points on a single line are collinear, we cannot choose all three points from the same line. Therefore, we have two cases to consider:
1. Choose 2 points from line AB and 1 point from line AC.
2. Choose 2 points from line AC and 1 point from line AB.

### Step 3: Calculate the Number of Combinations for Each Case
**Case 1: 2 points from AB and 1 point from AC**
- The number of ways to choose 2 points from 10 points on line AB is given by the combination formula \( \binom{n}{r} \), which is \( \binom{10}{2} \).
- The number of ways to choose 1 point from 8 points on line AC is \( \binom{8}{1} \).

Calculating these:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]
\[ \binom{8}{1} = 8 \]
So, the total combinations for this case:
\[ \text{Total for Case 1} = \binom{10}{2} \times \binom{8}{1} = 45 \times 8 = 360 \]

**Case 2: 2 points from AC and 1 point from AB**
- The number of ways to choose 2 points from 8 points on line AC is \( \binom{8}{2} \).
- The number of ways to choose 1 point from 10 points on line AB is \( \binom{10}{1} \).

Calculating these:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
\[ \binom{10}{1} = 10 \]
So, the total combinations for this case:
\[ \text{Total for Case 2} = \binom{8}{2} \times \binom{10}{1} = 28 \times 10 = 280 \]

### Step 4: Add the Two Cases Together
Now, we add the total combinations from both cases to get the final answer:
\[ \text{Total Triangles} = \text{Total for Case 1} + \text{Total for Case 2} = 360 + 280 = 640 \]

### Final Answer
The total number of triangles that can be formed is **640**.

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