If `20` lines are drawn in a plane such that no two of them are parallel and so three are concurrent, in how many points will they intersect each other?

Answer

To solve the problem of how many points 20 lines will intersect each other under the given conditions (no two lines are parallel and no three lines are concurrent), we can follow these steps:

### Step 1: Understand the conditions
We have 20 lines in a plane. The conditions state that:
- No two lines are parallel, meaning every pair of lines will intersect.
- No three lines are concurrent, meaning that no single point can be the intersection of more than two lines.

### Step 2: Determine the number of intersection points
Since no two lines are parallel and no three lines are concurrent, every pair of lines will intersect at a unique point.

To find the total number of intersection points formed by these lines, we need to calculate the number of ways to choose 2 lines from the 20 lines. This can be done using the combination formula:

\[ \text{Number of intersection points} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Where:
- \( n \) is the total number of lines (20 in this case),
- \( r \) is the number of lines to choose (2 for intersection).

### Step 3: Apply the combination formula
Substituting the values into the combination formula:

\[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20!}{2! \cdot 18!} \]

### Step 4: Simplify the expression
Now, we can simplify the expression:

\[ \binom{20}{2} = \frac{20 \times 19 \times 18!}{2! \times 18!} \]

The \( 18! \) cancels out:

\[ = \frac{20 \times 19}{2!} \]

Since \( 2! = 2 \):

\[ = \frac{20 \times 19}{2} = \frac{380}{2} = 190 \]

### Conclusion
Thus, the total number of points at which the 20 lines intersect each other is **190**.

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