permutations and combination
How many circles can be drawn passing through 3 points out of 6 given points, when one and only one set of 4 of these points are concoction?
Prove the following results
1.) By considering suitable selections
2.) By considering binomial expansion like (1+×)^k . Thanks in advance
Prove the following results
1.) By considering suitable selections
2.) By considering binomial expansion like (1+×)^k . Thanks in advance
Original post by Fondja Jordan
How many circles can be drawn passing through 3 points out of 6 given points, when one and only one set of 4 of these points are concoction?
Prove the following results
1.) By considering suitable selections
2.) By considering binomial expansion like (1+×)^k . Thanks in advance
Prove the following results
1.) By considering suitable selections
2.) By considering binomial expansion like (1+×)^k . Thanks in advance
I've not come across the term "concoction" in relation to mathematics before; what you do mean by it?
Original post by ghostwalker
I've not come across the term "concoction" in relation to mathematics before; what you do mean by it?
Sorry I meant "concyclic"
Original post by Fondja Jordan
Sorry I meant "concyclic"
OK, so we have four on a circle, and two other points ramdonly spaced.
Any thoughts on how you might proceed?
No I'm still thinking
Original post by Fondja Jordan
No I'm still thinking
OK, a couple of ideas spring to mind.
I'm assuming we want exactly three points on our circle. Also any three points, which aren't collinear, will define a circle.
a)
Forgetting about the fact that some points are concyclic for the moment, and consider 6 random points. How many circles can there be? Now subtract the number of combinations that arise from the concyclic set, since those circles will have 4 points on them, and not three.
b)
We could break it down into two cases.
Circles made from two of the points in the concyclic set, plus one other.
Circles made from one point in the concyclic set, plus two others.
We can't have circles where there are 3 points or 0 points from the concyclic set - why?
Edit: Not sure how the binomial expansion fits into this question.
(edited 7 years ago)
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