25 lines are drawn in a plane. Such that no two of them are parallel and no three of them are concurrent. The number of points in which these lines intersect, is:

Options

A. 300
B. 315
C. 325
D. 450

Answer

To find the number of intersection points formed by 25 lines drawn in a plane, where no two lines are parallel and no three lines are concurrent, we can follow these steps:

### Step 1: Understanding the Problem
We have 25 lines, and we need to determine how many unique intersection points can be formed. Since no two lines are parallel, every pair of lines will intersect at a unique point. Additionally, since no three lines are concurrent, each intersection point will be formed by exactly two lines.

### Step 2: Choosing Pairs of Lines
To find the number of intersection points, we need to calculate how many ways we can choose 2 lines from the 25 lines. The formula for choosing 2 items from n items is given by the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

In our case, \( n = 25 \) and \( r = 2 \).

### Step 3: Applying the Combination Formula
Substituting the values into the combination formula:

\[ \binom{25}{2} = \frac{25!}{2!(25-2)!} = \frac{25!}{2! \cdot 23!} \]

### Step 4: Simplifying the Expression
We can simplify this expression:

\[ \binom{25}{2} = \frac{25 \times 24}{2 \times 1} = \frac{600}{2} = 300 \]

### Step 5: Conclusion
Thus, the number of intersection points formed by the 25 lines is:

\[ \text{Number of intersection points} = 300 \]

### Final Answer
The number of points in which these lines intersect is **300**.

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