There are n points on a circle that are pairwise connected by a chord in the circle. What is the maximum and the minimum number of points within the circle that are intersections of the chords?
[Math] Finding the number of intersections of chords within a circle
combinatoricsgeometry
Related Solutions
I do not think there is a better approach to find the number of parts (part ii) comparing to the Euler's formula. But to use it one should first find the number of vertices and edges, and for this task I do not see how the Euler's formula can help.
For simplicity we disregard the circle arcs and will consider only the graph consisting of line segments connecting the points.
Let $s$ be a segment connecting two points of the circle. The points split the circle in two parts. Let the number of points in one of the parts be $k$. Then the number of points in the other part is $(n-k-2)$. Every segment which connects a point lying in one part with a point lying in the other part will intersect the segment $s$, whereas any segment connecting two points lying in the same part will not intersect $s$. Therefore the overall number of intersection points of the segment $s$ with other segments is $k(n-2-k)$ and the overall number of intersection points on all segments drawn from each of the considered circle points is: $$ v''_n=\sum_{k=0}^{n-2}k(n-2-k)=\frac{(n-1)(n-2)(n-3)}6=\binom{n-1}3\tag1 $$ and the overall number of the intersection points inside the circle is $$ v'_n=\frac n4 v''_n=\binom n4,\tag2 $$ where factor 4 in the denominator accounts for the fact that while summing over all $n$ circle points every internal intersection point is counted 4 times.
To compute the overall number of vertices we shall add the number of the circle points to the obtained result: $$ v_n=v'_n+n=\frac{n(n+1)(n^2-7n+18)}{24}.\tag3 $$
The number of edges can be computed similarly. Observe that the segment $s$ is split by $n_s$ internal intersection points into $n_s+1$ edges therefore: $$ e''_n=\sum_{k=0}^{n-2}[k(n-2-k)+1]=\binom{n-1}3+\binom{n-1}1=\frac{(n-1)(n^2-5n+12)}6,\tag4 $$ and the overall number of the edges inside the circle is $$ e'_n=\frac n2 e''_n=\frac{n(n-1)(n^2-5n+12)}{12},\tag5 $$ where factor 2 in the denominator accounts for the fact that while summing over all $n$ circle points every chord is counted 2 times.
To compute the overall number of edges we shall add the number of arcs connecting the circle points to the obtained result: $$ e_n=e'_n+n=\frac{n^2(n^2-6n+17)}{12}.\tag6 $$ Finally we use the Euler's formula for computing the number of parts inside the circle: $$ f_n=e_n-v_n+1=\frac{n^4-6n^3+23n^2-18n+24}{24}.\tag7 $$
And indeed the latter formula describes - as it should - the OEIS sequence A000127.
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.
Best Answer
To form a pair of chords that intersect, you need 4 points. Hence, that can be done in nC4 ways. There is only way to join the 4 points with two chords and form an intersection. Hence, the final answer is nC4.