In a plane, there are 15 distinct straight lines with no two parallel and no three meeting at a single point. What is the maximum possible number of intersection points of these lines?

Introduction / Context:
This question comes under the topic of combinatorics and geometry, often found in “Problems on Numbers” or counting principles. It asks for the maximum number of intersection points that 15 straight lines can form in a plane, under the conditions that no two lines are parallel and no three lines are concurrent (that is, they do not all meet at the same point). Understanding this helps in many counting and arrangement problems in competitive exams.

Given Data / Assumptions:

• There are 15 distinct straight lines drawn on a plane.
• No two lines are parallel.
• No three lines pass through the same intersection point.
• We must find the maximum number of intersection points formed by these lines.

Concept / Approach:
When no two lines are parallel, every pair of lines will intersect. When no three are concurrent, each pair intersects at a unique point. Therefore, the problem reduces to counting how many distinct pairs of lines we can form from 15 lines. The number of ways to choose 2 items from n items is given by the combination formula nC2 = n * (n − 1) / 2. This gives the maximum number of intersection points, because each pair of lines meets exactly once.

Step-by-Step Solution:
Step 1: Interpret the conditions. Since no two lines are parallel, every pair of lines intersects somewhere in the plane. Step 2: Because no three lines meet at a single point, each pair of lines produces a distinct intersection point, with no overlapping intersections. Step 3: The total number of intersection points is therefore equal to the number of pairs of lines. Step 4: Use the combination formula to find the number of pairs: 15C2 = 15 * 14 / 2. Step 5: Compute 15 * 14 = 210, and then divide by 2 to get 210 / 2 = 105. Step 6: Thus, the maximum possible number of intersection points is 105.

Verification / Alternative check:
You can verify the pattern with smaller numbers. For example, with 2 lines, you get 1 intersection point, and 2C2 = 1. With 3 lines in general position (no two parallel, no three concurrent), each pair intersects, giving 3C2 = 3 points. For 4 lines, you get 4C2 = 6 intersection points. The results match intuitive drawings. Extending this pattern to 15 lines gives 15C2 = 105, confirming that the formula works consistently and the reasoning is correct.

Why Other Options Are Wrong:
Option 55: This equals 11C2 and would be correct for 11 lines, not for 15 lines.
Option 215 and 148: These numbers do not match any direct combination count for 15 lines under the given conditions and are too large or inconsistent.
Option 90: This equals 10 * 9, which is unrelated to the required combination 15C2 in this context.

Common Pitfalls:
Students sometimes confuse combinations with permutations and may try formulas like nP2 instead of nC2. Another common error is forgetting the conditions: if some lines were parallel or three lines concurrent, the count would be smaller, but the question explicitly asks for the maximum, which assumes the general position described. Always check that the problem is asking for the maximum or minimum before deciding on the arrangement and the counting method.

Final Answer:
The maximum number of intersection points formed by 15 such lines is 105.