11 answer nC2
2 answer mc2 x nc2 (= to total number of quadrilateral)
491) 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
11) 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
·2010-11-09 14:32:55
i do not quite agree with u swordfish!
4 pts are not collinear! :D
well lets see wat jangra has to say! [1]
1ans is n/8(n-1)(n-2)(n-3)(n-4) i m quit close but not exactl.
23subhomoy 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
·2010-11-10 10:57:05
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
·2010-11-10 10:57:42
@Swordfish : now i somewhat agree with u! :D