if 20 straight line be drawn in a plane , no two of them being parallel and no three them concurrent , how many points of intersection will there be ?

Options

A. 95
B. 380
C. 400
D. 190

Answer

To find the number of points of intersection formed by 20 straight lines in a plane, where no two lines are parallel and no three lines are concurrent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 20 straight lines.
- No two lines are parallel, meaning every pair of lines will intersect.
- No three lines are concurrent, meaning that each intersection point is formed by exactly two lines.

2. **Finding the Number of Intersections**:
- The number of intersection points formed by the lines can be calculated using combinations. Specifically, we need to choose 2 lines out of the 20 to find the intersection point.
- The formula for combinations is given by:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
- Here, \( n \) is the total number of lines (20), and \( r \) is the number of lines we are choosing (2).

3. **Applying the Formula**:
- We need to calculate \( C(20, 2) \):
\[ C(20, 2) = \frac{20!}{2!(20-2)!} = \frac{20!}{2! \cdot 18!} \]
- Simplifying this, we can cancel out the \( 18! \) in the numerator and denominator:
\[ C(20, 2) = \frac{20 \times 19}{2 \times 1} = \frac{380}{2} = 190 \]

4. **Conclusion**:
- Therefore, the total number of points of intersection formed by the 20 lines is **190**.

### Final Answer:
The number of points of intersection is **190**.