# 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.

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 = $7-4$ = $3$

• Where is the question? – Marc van Leeuwen Mar 5 '13 at 6:00
• Rajesh: It would be good for you to consider accepting answers when they are helpful. You can accept one answer per question, and you can do so by clicking on the $\checkmark$ to the left of the answer you'd like to accept. Plus, bonus: you get two reputation points each time you accept an answer! – Namaste Mar 7 '13 at 9:08

• If you don't mind, may I ask you to please elaborate further. I mean you are talking about three subtractions of two as done by OP. But I am failing to understand the reasoning behind these subtractions. Moreover, what is $^6C_2$ for? – Ramit Sep 13 '13 at 10:43
• @Ramit: I don't understand OP's calculation, either. $^6C_2$ is the number of ways to choose two of the six points, which would be the total number of lines if no three points were collinear. I think the three $-2$s were to account for the ones missing because of collinearity, in which case there should be $4$ of them and the result $7$ is correct. – Ross Millikan Sep 13 '13 at 13:15