# Two questions on Perm com

1. Let there be m and n points on two straight lines in a plane(not necessarilly parallel). Join all the points in every way possible. Prove that other than the m+n points lying on the two lines, the number of intersectionss is mn(m-1)(n-1)/2 2.

Let there be 2n lines on a plane of lengths 1,2,3,.....2n. Prove that the number of possible triangles from these lines is n(n-1)(4n-5)/6

10/10 I would do the same.

But Rajorshi must have a better way of solving it.

dudes !!! i dont have the solution . still working on it ! i am not that a sadist . I saw this problem in Chryshtal's algebra perhaps in the same or higher league the Hall and Knight. But even if we cant solve it its good to remember the results!

sorry guys the first question the proof should be for mn(m-1)(n-1)/4

you can check that for m = 2 , n =2

i got that one! It was pretty easy I must say.

Solution: Just to keep things simple, I will disregard the intersection points on the line.

And its better to have a pencil and paper with you while you read through the solution 'cos i am trying my best to explain it in words and without a figure.

Lets have a notation.

Let the m points on the 1 st line be m1, m2 , m3 ..... mm.

let the line drawn from mi to nj be called mi->nj.

lets start with m to the 1st line and 1 point on 2 nd line. clearly the number of intersections is zero. Let this be Set1

now lets add another point on the 2nd line and start joining the new point to the m points on the 1st line. We will call this Set2

You will notice the n2->m1 line intersects all Set1 lines except m1->n1 ,

so the number of intersection points is m-1.

n2->m2 line intersects all Set1 lines except m1->n1 and m2->n1,

so the number of intersection points is m-2.

and so on. Also n2->mm line does not intersect any Set1 and Set2 lines,

so the number of intersection points is 0.

So the total number of points of intersection for m and 2 points = (m-1)m/2

Now Lets add another point n3 and repeat the process. Till now we have drawn

m*2 lines

Here we will note.

You will notice the n3->m1 line intersects all Set1 and Set2 lines except

m1->n1 and m1->n2

so the number of NEW intersection points is 2m-2 = 2(m-1).

n2->m2 line intersects all Set1 and Set2 lines except m1->n1 , m1->n2 ,

m1->n2, m2->n2

so the number of intersection points is 2m-4 = 2(m-2).

and so on. Also n2->mm line does not intersect any Set1 lines,

so the number of intersection points is 0.

So the total number of points of intersection for m and 3 points = 2(m-1)m/2

I HOPE YOU ARE SEEING THE PATTERN NOW .

Similarly, the total number of NEW points of intersection

for m and 4 points = 3(m-1)m/2

Thus the total number of new points of intresection for m and i points would be

(i-1)(m-1)m/2

thus the TOTAL NUMBET OF POINTS would be : (m(m-1)/2 )(1 + 2 + .... + n-2 + n-1)

= mn(m-1)(n-1)/4

Thanks Rajorshi for explaining it with such minute details.

Initially I approached like that but then thought there might be a easier method with permutation and combination

there could be but would need someone one of better IQ than me to figure that out !!!

The question number 1 is correct i.e, the number of intersections is mn(m-1)(n-1)/2.

If we take 2 points out of the m points (lets say m1 and m2) and then take 2 points out of the n points (lets say n1 and n2) , we get 2 points of intersection.

The first point is given by intersection of two lines formed by joining m1 to n1 and m2 to n2. The second point is given by intersection of two lines formed by joining m1 to n2 and m2 to n1.

Hence , the total no. of intersection points = (m C 2) * (n C 2) * 2

= m(m-1)/2 * n(n-1)/2 * 2 = mn(m-1)(n-1)/2.

lets say we have a quadrilateral ! that has 2 points on one line and 2 points on the other. the number of intersections other than the 4 points themselves by joining all the points in all possible ways is the THE POINT OF INTERSECTION OF THE DIAGONALS.The question says JOINING THE POINTS IN ALL POSSBLE WAYS. it doesn't say extending the lines beyond the points. So that way mn(m-1)(n-1)/2 does't seem to be the answer.

hii...um nt brainy enuf to solv this kinoff problm vid da approach that v all r luking here

bt yeps i cn solve dis if i wud hv got this problm in CAT...ofcourse u cn also solve dat rajj...

i wud hv solve dat...by taking 2 pts on 1 line and 3 pts on 2nd line n then join all thosee pts...

frm dis i'l calculate the values of m,n, m+n, mn and by putting it into the options...i'l get the ansr...

sumtimes u got 2 follow sum othr crooky methods....