New lines from given lines

Probability Level 2

There are 'n' straight lines in a plane , no two of which are parallel , and no three pass through a same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is

Note:- A point of intersection is not collinear with any other points of intersections which do not line on the two lines whose point of intersection we are considering.

n ( n − 1 ) ( n − 2 ) ( n − 3 ) 6 \frac { n(n-1)(n-2)(n-3) }{ 6 } n ( n − 1 ) ( n − 2 ) ( n − 3 ) 8 \frac { n(n-1)(n-2)(n-3) }{ 8 } n ( n − 1 ) ( n − 2 ) 8 \frac { n(n-1)(n-2) }{ 8 } n ( n − 1 ) ( n − 3 ) 6 \frac { n(n-1)(n-3) }{ 6 }

Mihir Mistry by Mihir Mistry , 24, India

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1 solution

Mihir Mistry
Sep 14, 2014

1) There are n n lines, so we get n C 2 ^{ n }{ C }_{ 2 } points of intersection.

2) A line will have ( n − 1 ) (n-1) points of intersection.(obviously they will be co-linear )

Now,select any one point of intersection It will be co-linear with 2 sets of ( n − 2 ) (n-2) points lying on the two lines whose point of intersection we have selected. And joining the selected point with these points will give us no fresh line. Hence the selected point is joined with n C 2 − { 2 ( n − 2 ) + 1 } ^{ n }{ C }_{ 2 }-\{ 2(n-2)+1\} ....(subtracting the two set of co-linear points and the selected point from total points)

solving it we get ( n − 2 ) ( n − 3 ) 2 \frac { (n-2)(n-3) }{ 2 }

we have such n C 2 ^{ n }{ C }_{ 2 } points of intersection so multiplying with it, we get n ( n − 1 ) ( n − 2 ) ( n − 3 ) 4 \frac { n(n-1)(n-2)(n-3) }{ 4 }

But wait this is not the answer,on multiplying with n C 2 ^{ n }{ C }_{ 2 } , we see that each line will be repeated twice , hence the above result should be divided by 2 and we get the answer n ( n − 1 ) ( n − 2 ) ( n − 3 ) 8 \frac { n(n-1)(n-2)(n-3) }{ 8 }

It is my first problem and solution so sorry if I am not clear....

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I agree that if we were to simply take "pairs of points of intersections from distinct lines", then the answer is ( n 2 ) × ( n − 2 2 ) { n \choose 2 } \times { n - 2 \choose 2 } .

My concern is that we could have 3 points on intersection which are collinear, and hence only contribute 1 "fresh line", instead of 3 "fresh lines" like you would want.

Can you clarify this in the problem statement?

Calvin Lin Staff - 6 years, 9 months ago

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Calvin i guess you want to say that a point of intersection is not collinear with any other points of intersections which do not line on the two lines whose point of intersection we are considering. If so you may please make the suitable corrections in the problem.

Mihir Mistry - 6 years, 9 months ago

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