There are p,q,r points on three parallel lines `L_1, L_2` and `L_3` all of which lie in the plane. The number of triangles which can be formed with vertices at their points is

Options

A. `"^(p+q+r)C_3`
B. `""^(p+q+r) C_3-""^p C_3- ""^qC_3 -"^r C_3`
C. `""^p C_3+""^qC_3+"^r C_3`
D. none of these

Answer

To solve the problem of finding the number of triangles that can be formed with vertices at points on three parallel lines \( L_1, L_2, \) and \( L_3 \), where there are \( p, q, \) and \( r \) points on these lines respectively, we can follow these steps:

### Step 1: Understand the Total Points
We have:
- \( p \) points on line \( L_1 \)
- \( q \) points on line \( L_2 \)
- \( r \) points on line \( L_3 \)

The total number of points is \( p + q + r \).

### Step 2: Calculate Total Combinations of Points
To form a triangle, we need to select 3 points. The total number of ways to select 3 points from \( p + q + r \) points is given by the combination formula:
\[ \binom{p + q + r}{3} \]

### Step 3: Exclude Collinear Points
Since all points on the same line are collinear, they cannot form a triangle. Therefore, we need to exclude combinations of points that are all from the same line:
- From \( L_1 \): The number of ways to choose 3 points from \( p \) points is \( \binom{p}{3} \).
- From \( L_2 \): The number of ways to choose 3 points from \( q \) points is \( \binom{q}{3} \).
- From \( L_3 \): The number of ways to choose 3 points from \( r \) points is \( \binom{r}{3} \).

### Step 4: Final Calculation
Now, we can find the total number of triangles that can be formed by subtracting the collinear combinations from the total combinations:
\[ \text{Number of triangles} = \binom{p + q + r}{3} - \left( \binom{p}{3} + \binom{q}{3} + \binom{r}{3} \right) \]

### Conclusion
The final expression for the number of triangles that can be formed with vertices at the given points is:
\[ \text{Number of triangles} = \binom{p + q + r}{3} - \binom{p}{3} - \binom{q}{3} - \binom{r}{3} \]