permutation and combination- A Das Gupta

1.There are n≥4 points situated on the plane such that no two of the lines joining them are parallel and no three of lines joining them are concurrent except at the given points.Find the number of points of intersection, other then the given points, of the lines obtained by joining them.
2.Straight lines are drawn by joining m points on a straight line to n points on another line.Then excluding the given points, prove that the lines drawn will intersect at 1/2 mn(m-1)(n-1) points if no two lines drawn are parallel and no three lines are concurrent.

9 Answers

1
ayush_2008 ·

1 answer nC2
2 answer mc2 x nc2 (= to total number of quadrilateral)

1
jangra28192manoj jangra ·

totally wrong.no way near

49
Subhomoy Bakshi ·

1) if we chose 4 arbitrary points from the n points, we will be able to visualise the formation of a quadrilateral!!

now in this quadrilateral, the points where all the possible lines intersect are...the 4 original points and one where the diagonals meet..

so on selecting 4 points, we get 1 new point where the lines drawn meet..

thus the probable answer is nC4

1
swordfish ·

1) Out of n points, two points are needed for a straight line.
Hence there are nC2 number of total lines, which infact represent the points of intersections.
Out of n points, 4 points line on the same line. So we need to subtract all lines passing through these. Therefore net points of intersection=
nC2 - nC4.
Wait! there is one line that passes through all these points and 4 points of intersections. So there are in total
nC2 - nC4 + 4 points of intersection.

I think this should be the correct answer.

49
Subhomoy Bakshi ·

i do not quite agree with u swordfish!

4 pts are not collinear! :D

well lets see wat jangra has to say! [1]

1
jangra28192manoj jangra ·

ans is n/8(n-1)(n-2)(n-3)(n-4) i m quit close but not exactl.

23
qwerty ·

subhomoy u missed somthing in the first question ,
for a new pt u need 2 lines to intersect, and those 2 lines can be formed using 4 pts , agreed , BUT the total no of lines that these 4 pts can make , cr8 in all 3 new pts of intersection ,

hence for first i go with 3 x nC4

49
Subhomoy Bakshi ·

hmm....i got it now...

if the question had been an n sided polygon and we had to find the no of internal points in which the the diagonals intersect..then my anwer would have been the answer!!! :)

49
Subhomoy Bakshi ·

@Swordfish : now i somewhat agree with u! :D

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