Permutation & Combinations

Answer: 300.

Now, I am not sure if there is any formula for this problem. Here's how I did it.

I took a much smaller sample space. 4 lines.

if no 2 lines are parallel and no 3 lines are concurrent then every line is intersected at 3 points.
so for 4 lines, total points of intersection are 4 * 3. But here we count each point twice and in fact there are only 6 points of intersection.
So, for 4 lines, total points of intersection are = (4 * 3)/2 = 6

likewise for 25 lines, total points of intersection are = (25 * 24) / 2 = 25 * 12 = 300

Hope this helps.

alaynaik wrote: Que: If 25 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then in how many points do they intersect?
Thanks

Hi,
A point of intersection is where two lines intersect. So, for every two lines, there is a point of intersection. It is nothing but choosing any 2 lines from the 'n' lines.
The number of ways is nC2.
In this case, it is 25C2 = 25*24/2 = 300.