Four lines are drawn on a plane with no two parallel and no three concurrent. Lines are drawn joining the points of intersection of these four lines. How many new lines are there now?
$4C_{2}=6$
$6C_{2}=15 -2 -2 -2=7$
New lines obtained = $7-4$ = $3$
Best Answer
You start with four lines and six points of intersection. If you draw the figure, there are three of the six points on each line, so the six points do not determine fifteen lines. You don't explain the reasoning behind the three subtractions of two. Presumably you are claiming that the end figure has seven lines in it. I believe that is true, but it needs some argument. I would say that each point has two lines through it which connect it to four other points, so you can construct one new line through each point, but have counted each new line twice, giving three.
A figure is below. The light lines are the original four, the dots are the points, and the three heavy lines are the new ones.