# Number of points of intersection between lines.

Lines $L_1,L_2,...,L_{100}$ are distinct .All lines $L_{4n}$ ,$n$ a positive integer, are parallel to each other. All lines $L_{4n-3}$, $n$ a positive integer,pass through a given point $A$. Find the maximum number of point of intersection of pairs of lines from the complete set ${L_1,L_2,...,L_{100} }$.

Options:

$a) 4350$

$b) 4351$

$c) 4900$

$d) 4901$

My attempt:

I've calculated the number of points of intersection of pairs of lines with none of them parallel to some other line,so I had $\dbinom {100}{2}$.

From this I have subtracted the pairs of lines which are parallel to each other,yelding $\dbinom {100}{2} - \dbinom {25}{2} = 4650$ .

But this is wrong if I look at the options given.Where's my mystake ?

Each non-parallel and parallel line will have $1$ point of intersection,isn't it ?

• Altogether, $100$ lines have at most ${100\choose2}=4950$ intersection points.
• You lose all the ${25\choose2}=300$ intersection points of the $25$ parallel lines.
• You lose all but one of the ${25\choose2}=300$ intersection points of the lines through point $A$.
This yields a maximum number of $4950-300-299=4351$ intersection points.
• Why do I lose $300$ points of intersection of the lines through $A$ ?We count all these intersections as $1$ since they pass through the same point ? – Nameless Nov 20 '15 at 10:37