Hi I'm working on my scholarship exam practice but I got stuck at my last question, could you please have a look? I'll write every part of question so you know the big picture. I have marked the problematic question in bold below.
This question only assumes knowledge in high school or pre-university level.
Draw a circle $C$ on a plane. Put $n$ distinct points on the circumference of $C$. Join any two of the $n$ points with a chord. Suppose that no three chords intersect at any one point inside $C$. Let $c_n$ be the number of such chords, $i_n$ that of intersections of the chords inside $C$, and $r_n$ that of regions in $C$ bounded by an arc and/or some chords. Fill in the blanks with the answers to the following questions.
(1) Find $c_1, c_2, c_3, c_4, c_5, c_6$ and write their values in this order.
The answer of this part checked with answer key is $0, 1, 3, 6, 10,$ and $15$ by drawing 6 different circles with n dots from 1 to 6 each circle, connecting each dot and counting number of chords.
(2) Find $i_1, i_2, i_3, i_4, i_5, i_6$ and write their values in this order.
The answer of this part checked with answer key is $0, 0, 0, 1, 5,$ and $15$ by counting intersections of chords.
(3) Find $r_1, r_2, r_3, r_4, r_5, r_6$ and write their values in this order.
The answer of this part checked with answer key is $1, 2, 4, 8, 16,$ and $31$ by counting the regions.
(4) Express $r_n$ with binomial coefficients in terms of $n$. (For binomial coefficients ${m}\choose{k}$, note that ${m}\choose{k}$$= 0$ if $m<k$.)
I got stuck at this question since I got nowhere near the answer key.
My approach is below:
${n}\choose{k}$$=\frac{n!}{(n-k)!\cdot k!}$
Pick $n=2$ and $n=3$ and substitute the values from (3) into the equation,
${2}\choose{k}$$=\frac{2!}{(2-k)!\cdot k!}=2$
${3}\choose{k}$$=\frac{3!}{(3-k)!\cdot k!}=4$
$\frac{2!}{(2-k)!\cdot k!\cdot2}=\frac{3!}{(3-k)!\cdot k!\cdot4}$
And $k$ will be weirdly equal to $\frac{3}{2}$ leading to binomial coefficient ${n}\choose{3/2}$, which is incorrect.
The answer key provided is $1 +$${n}\choose{2}$$+$${n}\choose{4}$.
How can we get to this answer? And, how is my method faulty?
