Help with Combinatorics - Part 2 (Prophet Sir n All)

Sir, I need help with those special type of questions like -

1. Find the maximum no. of planes in which n circles can divide a plane?

2. Find the maximum no. of point of intersection of n circles and n straight lines?

3. Total no. of Integral Solutions of xyz = 24 ?

\textup{ONE DOUBT::}

4 points out of 8 in a plane are collinear. Number of different quadrilaterals that can be formed by joining them is -
a) 56 b) 53 c) 76 d) 60

Soln: Sir I thought it should be 8C4 - 4C4 - 4C3 =65

Help me!

4 Answers

62
Lokesh Verma ·

I will give you hints.. see if you can work on those..

2) Two circles intersect in atmost 2 points (except if they are identical)

2 straight lines in atmost 1 point

and a circle and a straight line in 2 points...

Now try this one..

3) it is a the same as 2^3*3
so how many ways can you distribute the powers of 2 in a, b and c such that they add up to 3

same with powers of 3

now there is one small thing here a, b and c can be +ve or -ve

so you have some - signs to take care of... can you :)

1
varun.tinkle ·

in the last answer

i dont think there should be a reason for 4c3

71
Vivek @ Born this Way ·

Okay! 1st done, 2nd done. Doubt still in 3rd.

Now for the last one, I found myself missing something which was-

8C4 - 4C4 - 4C3 * 4C1
= 53

One of the options matches.

Now again if i think the other way, Say

Coll. Points : 1 2 0
Non. Coll Points : 3 2 4

These will be the cases where a quadrilateral can be formed.
WHich gives 4C1*4C3 + 4C2*4C2 + 4C0*4C4 = 53

Either one gives same I think!!

71
Vivek @ Born this Way ·

I post my solution for xyz = 24

Since, xyz = 24 = 23 * 31

Hence, We have α+β+γ = 3 and α+β+γ = 1 where α,β.γ ≥0

Therefore, No. of Solutions = 5C2*3C2 = 30 [By Multinomial]

But we can assign two negatives to any two of the three to have a positive solution.

Hence, Finally 30*3C2 = 90 ANS

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