The fiftieth question, given a positive integer $n \geq 3$, let $S$ be a finite set of points in the plane, $|S|=n$, with no two points collinear and no four points concyclic. If each point can be colored either red or white, such that there exists a circle with all points inside it being white and all points outside it being red, then this coloring is called "proper." How many proper coloring schemes are there?

Given a positive integer $n \geq 3$, let $S$ be a finite set of points in the plane, $|S|=n$, with no two points collinear and no four points concyclic. If each point can be colored either red or white, such that there exists a circle with all points inside it being white and all points outside it being red, then this coloring is called "proper." How many proper coloring schemes are there?
Fiftieth Question,
Solution: We analogize the inversion and polarity on a plane to define inversion and polarity in three-dimensional space. The following specifies that the inversion and polarity center is point $\mathrm{O}$. First, we provide several basic properties:

Property 1: A plane passing through point $\mathrm{O}$ is its own inverse; a plane not passing through $\mathrm{O}$ and a sphere passing through $\mathrm{O}$ are mutual inverses; a sphere not passing through $\mathrm{O}$ and another sphere not passing through $\mathrm{O}$ are mutual inverses.
Property 2: Four points in space are concyclic if and only if their inverses are concyclic.
Property 3: For two points $\mathrm{A}$ and $\mathrm{B}$ in space, $\mathrm{A}$ and $\mathrm{O}$ are on the same side of $\mathrm{B}$'s polar plane if and only if $\mathrm{B}$ and $\mathrm{O}$ are on the same side of $\mathrm{A}$'s polar plane.
We first prove two lemmas.
Lemma 1: In space, $n$ planes that are not parallel and no three of which are collinear can divide the space into $f(n) = C_{n}^{3} + C_{n}^{2} + C_{n}^{1} + C_{n}^{0}$ regions.
Proof of Lemma 1: First, it is well-known that $n$ lines on a plane can divide the plane into $\frac{n(n+1)}{2} + 1$ parts.
Clearly, $f(1) = 2$, $f(2) = 4$, $f(3) = 8$. When there are $n$ planes, the space can be divided into $f(n)$ regions. Adding the $(n+1)$-th plane, the other $n$ planes intersect on this new plane, forming $n$ lines. These $n$ lines can divide the new plane into $\frac{n(n+1)}{2} + 1$ parts. This indicates that adding the $(n+1)$-th plane increases the number of regions by $\frac{n(n+1)}{2} + 1$, thus
$$
f(n+1) - f(n) = \frac{n(n+1)}{2} + 1
$$

It is easy to solve that $f(n) = C_{n}^{3} + C_{n}^{2} + C_{n}^{1} + C_{n}^{0}$.
Lemma 2: Given a set of $n+1$ points in three-dimensional space, where no three points are collinear and no four points are coplanar, a partition $P = A \cup B$ of $P$ is "good" if and only if there exists a plane $M$ such that $A$ and $B$ are on opposite sides of $M$. Let $f(n) = C_{n}^{3} + C_{n}^{2} + C_{n}^{1} + C_{n}^{0}$, then there are exactly $f(n)$ good partitions.

Proof of Lemma 2: Choose any point $O$ from $P$, and let $M$ and $N$ be two planes. Note that the partitions of $P$ by $M$ and $N$ are different if and only if there exists a point $A$ in $P$ such that $O$ and $A$ are on the same side of $M$ but on opposite sides of $N$, or $O = A$ is on the same side of $N$ but on opposite sides of $M$.

Using $O$ as the center of polarity, the other $n$ points in $P$ are polarized to $n$ planes. By induction, it is easy to see that these $n$ planes divide the space into $f(n)$ regions. According to Property 3, the partitions of $P$ by $M$ and $N$ are different if and only if there exists a point $A$ such that the polar points of $M$ and $N$ are on opposite sides of $A$'s polar plane, which is equivalent to the polar points being in different regions. Thus, the number of good partitions is the number of regions. The lemma is proved!
Now, let's return to the original problem.
Choose any point $O$ not on the given plane, and perform inversion with $O$ as the center. Let the inverse of point set $S$ be $S'$, and let $P = S' \cup \{0\}$.

According to Property 1 and Property 2, $S'$ lies on a sphere $C$ passing through $O$, and no four points in $S'$ are concyclic. Consider a plane $M$ that intersects $C$ but does not pass through any point in $S'$. Note that the intersection of $C$ and $M$ is a circle $T'$ not passing through $O$. Let $T$ be the inverse of $T'$, then the inverse of the points in $S'$ on the same side of $O$ as $T'$ lies inside $T$. This establishes a mapping from the original red and white coloring to the good partitions of $P$. According to the lemma, $P$ has exactly $f(n) = C_{n}^{3} + C_{n}^{2} + C_{n}^{1} + C_{n}^{0}$ good partitions.
In conclusion, the number of appropriate coloring schemes that satisfy the conditions is $C_{n}^{3} + C_{n}^{2} + C_{n}^{1} + C_{n}^{0}$.