1in the last answer
i dont think there should be a reason for 4c3
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!
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4 Answers
62I 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 :)
71Okay! 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!!
71I 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