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There are $p\ge6$ points given on circumference of a circle, and every two points are joined by a chord. Assume that no 3 chords are concurrent within the interior of the circle. We have to find the number of intersection points made by these chords inside the circle.


We count the ways to choose 4 points out of $p$ and each set of the chosen points contributes to a single intersection point inside the circle. So the answer is $$p\choose4$$ But I cannot understand the significance of the condition in bold. How will the answer change without this condition?

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    $\begingroup$ That assumption is there, because otherwise the answer would not be unique. If you have got the answer to this question, you should write it down with the question, so that we can point out, at which step of the answer have you implicitly used the assumption. $\endgroup$ – астон вілла олоф мэллбэрг Jul 11 '18 at 6:39
  • $\begingroup$ @астонвіллаолофмэллбэрг I don't understand how the answer will not be unique $\endgroup$ – naman jain Jul 11 '18 at 6:45
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    $\begingroup$ Think of $n=6$ and the configuration where the points are the vertices of a regular hexagon. That will have fewer intersection points than the generic case. $\endgroup$ – Lord Shark the Unknown Jul 11 '18 at 6:46
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    $\begingroup$ Each set of points contributes to a single intersection point inside the circle, correct. How do you know that two different sets of points always contribute different end points? If this were not true, then you would overcount the intersection points. The hypothesis ensures that this is true. $\endgroup$ – астон вілла олоф мэллбэрг Jul 11 '18 at 7:07
  • $\begingroup$ Same formula works for $p=4,5.$ $\endgroup$ – coffeemath Jul 11 '18 at 7:46
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Suppose you have three concurrent lines (i.e. all intersecting at a single point). Now perturb the position of the line endpoints slightly. Almost surely the lines no longer intersect a single point; instead you have three pairwise intersections and a tiny triangle where the single intersection point used to be. Therefore any time you have (exactly) three concurrent lines, you "lose" two intersection points from the number predicted by your formula.

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