3. Let $A$ be the set of any 20 points on the circumference of a circle. Joining any two points in $A$ produces one chord of this circle. Suppose every three such chords are not concurrent. Find the number of regions within the circle which are divided by all these chords.
3. Ans: 5036
Consider such a figure as a plane graph $G$. Then the answer is equal to the number of interior faces of $G$. The number of vertices in $G$ is $20+\binom{20}{4}$. The sum of all degrees is
$$
20 \times 21+4 \times\binom{ 20}{4}
$$
and so the number of edges in $G$ is
$$
10 \times 21+2 \times\binom{ 20}{4} .
$$

Hence the number of interior faces, by Euler's formula, is
$$
\binom{20}{4}+10 \times 21-20+1=5036 .
$$

Second Solution: Let $P_{n}$ be the number of such regions with $n$ points on the circumference. Then $P_{1}=1, P_{2}=2$ and in general, for $n \geq 2, P_{n+1}=P_{n}+n+\sum_{i=1}^{n-2} i(n-$ i). This can be obtained as follows. Suppose there are $n+1$ points $a_{0}, \ldots, a_{n}$ in that order on the circumference. The chords formed by $a_{1}, \ldots, a_{n}$ create $P_{n}$ regions. The chord $a_{0} a_{1}$ adds one region. The chord $a_{0} a_{1}$ adds $1+1 \times(n-2)$ regions as this chord intersects the existing chords in $1 \times(n-2)$ points. Similarly, the chord $a_{0} a_{2}$ adds $1+2 \times(n-3)$ regions, etc. From this, it is easy to show that
$$
\begin{aligned}
P_{n} & =1+\binom{n}{2}+\binom{n-2}{2}+2\binom{n-3}{2}+\cdots+(n-3)\binom{2}{2} \\
& =1+\binom{n}{2}+\binom{n-1}{3}+1\binom{n-3}{2}+\cdots+(n-4)\binom{2}{2} \\
& =\cdots=1+\binom{n}{2}+\binom{n-1}{3}+\binom{n-2}{3}+\cdots+\binom{3}{3}=1+\binom{n}{2}+\binom{n}{4}
\end{aligned}
$$

Thus $P_{20}=1+\binom{20}{2}+\binom{20}{4}=5036$.