I think you can do this with probability. Let's colour the points black and white and Let the probability that a point has colour black be x. Now look at any convex polygon $ P$. The probability that all its vertices are black is $ x^{a(P)}$ and that all the points outside it are white is $ (1 - x)^{b(P)}.$ The given summation is the probability that there exists a polygon with vertices black and all points outside it are white. But we know this is true (i.e. 1) as all we have to do is consider the convex hull of the black points. Now this is true for all $ x\in(0,1)$. But this is a polynomial with finite degree and the number of points at which it evaluates 1 is infinite. so, this is an identity and holds $ \forall x\in \mathbb{R}$ Moderator edit: See also the discussion of Lubell-Yamamoto-Meshalkin inequality.

watchcatno1 wrote: Let's colour the points black and white and Let the probability that a point has colour black be x. Now look at any convex polygon $ P$. The probability that all its vertices are black is $ x^{a(P)}$ and that all the points outside it are white is $ (1-x)^{b(P)}.$ Very very beautiful!! Quote: The given summation is the probability that there exists a polygon with vertices black and all points outside white. Or: The given summation is the expected number of polygons with black vertices and all points outside white. Since for every fixed coloring, there exists exactly one such polygon, the expected number over all such colorings equals 1.

Yeah that's right.

I know using the probabilistic method pioneered and championed by Erdös, Alon, Spencer is a very neat solution. Maybe somebody wants to discuss a different approach.

Let $ f(P)=x^{a(P)}(1-x)^{|S|-a(P)}$. By the binomial theorem, $ \sum _{P\in S} f(P)=1$ where $ P$ is any set of points in $ S$ not just convex polygons. For any convex polygon $ P$, the value of $ \sum _Q f(Q)$, where the sum is taken over all sets of polygons $ Q$ such that all points of $ Q$ are in or on $ P$ and $ P\in Q$ (i.e. all polygons with $ P$ as its convex hull), is $ \sum ^{|S|-b(P)}_{k=a(P)}\dbinom{|S|-a(P)-b(P)}{k-a(P)}x^{k}(1-x)^{|S|-k}=x^{a(P)}(1-x)^{b(P)}$. Since every set of points has exactly one convex hull, $ \sum _{P\in S} f(P)=\sum _{P}x^{a(P)}(1-x)^{b(P)}=1$, as desired.

A different approach using strong induction: For any finite set $T$ of points in a plane with no three collinear points, Let $P^*_T$ be the set of all convex polygons whose vertices are in $T$, $b_T(P)$ be the number of points in $T$ which lies outside $P$ We assume that $\sum_{P\in P^*_{T}}\left[x^{a(P)}(1-x)^{b_{T}(P)}\right]=1$ forall $T$ with $|T|<|S|$ We can check for base case to be true. Let $S$ be the universal set here, i.e. for any set $X\subseteq S$, we denote $X'=S\setminus X$ Let $C$ be the convex hull of $S$. For any finite non-empty set $A$ and any number $z,$ $\sum_{X\subseteq A}z^{|X|}=(1+z)^{|A|}\cdots (I)$ For $X\subseteq C$, $P^*_{X'}\subseteq P^*_S$ and $b_S(P)=b_{X'}(P)+|X|\ \forall P\in P^*_{X'}\cdots (II)$ For a fixed $P\in P^*_S$ such that $P\neq C$, $\sum_{X\subseteq C, P\subseteq X'}(-1)^{|X|}=0$ (sum ranging over all sets $X$ satisfying $X\subseteq C, P\subseteq X'$) this follows from $(I)$ with $A=C\cup P'$ and $z=-1$ Now, for $P\neq C$: \[0=x^{a(P)}(1-x)^{b_S(P)} \left[ \sum_{X\subseteq C, P\subseteq X'}(-1)^{|X|} \right]\] \[=\sum_{X\subseteq C, P\subseteq X'}\left[(-1)^{|X|}x^{a(P)}(1-x)^{b_S(P)}\right]\] For $P=C$, \[\sum_{X\subseteq C, P\subseteq X'}\left[(-1)^{|X|}x^{a(P)}(1-x)^{b_S(P)}\right]\] \[=\sum_{X=\phi}\left[(-1)^{|X|}x^{a(P)}(1-x)^{b_S(P)}\right]=(-1)^0x^{|C|}(1-x)^0=x^{|C|}\] Now, by varying $P$ we have: \[x^{|C|}\] \[=\sum_{P\in P^*_S}\sum_{X\subseteq C, P\subseteq X'}\left[(-1)^{|X|}x^{a(P)}(1-x)^{b_S(P)}\right]=\]\[\text{(from } (II) \text{ we have)}\] \[=\sum_{X\subseteq C}\sum_{P\in P^*_{X'}}\left[(-1)^{|X|}x^{a(P)}(1-x)^{b_{X'}(P)+|X|}\right]=\] \[=\sum_{X\subseteq C}(x-1)^{|X|}\sum_{P\in P^*_{X'}}\left[x^{a(P)}(1-x)^{b_{X'}(P)}\right]=\] \[=\sum_{X\subseteq C,X\neq \phi}(x-1)^{|X|}\sum_{P\in P^*_{X'}}\left[x^{a(P)}(1-x)^{b_{X'}(P)}\right]+\sum_{P\in P^*_{S}}\left[x^{a(P)}(1-x)^{b_{S}(P)}\right]=\]\[\text{(from induction hypothesis}\sum_{P\in P^*_{X'}}\left[x^{a(P)}(1-x)^{b_{X'}(P)}\right]=1\ \forall |X'|<|S|\text{)}\] \[\sum_{X\subseteq C,X\neq \phi}(x-1)^{|X|}\cdot 1+\sum_{P\in P^*_{S}}\left[x^{a(P)}(1-x)^{b_{S}(P)}\right]\] So \[\sum_{P\in P^*_{S}}\left[x^{a(P)}(1-x)^{b_{S}(P)}\right]\] \[=x^{|C|}-\sum_{X\subseteq C,X\neq \phi}(x-1)^{|X|}=x^{|C|}+1-\sum_{X\subseteq C}(x-1)^{|X|}=x^{|C|}+1-[1+(x-1)]^{|C|}=\]\[\text{(follows from }(I)\text{ with }A=C,z=(x-1)\text{)}\] \[1\] QED

We use strong induction on $|S|$, where the base case $|S|=0$ is clear. Consider the convex hull of $S$, and say it has $n$ points $Q_1,Q_2,\ldots,Q_n$. Then by the inductive hypothesis, (letting $S'=S-\{Q_{i_1},Q_{i_2},\ldots,Q_{i_k}\}$) \[f(S')=\sum_{P\in S'}x^{a(P)}(1-x)^{b_{S'}(P)}=1,\]so using PIE to sum up $x^{a(P)}(1-x)^{b_{S}(P)}$ over all convex polygons $P\in S$ excluding at least one point on the convex hull $Q_1Q_2\cdots Q_n$ (note that the $k$ points excluded are outside $P$), we obtain \[f(S)-x^n(1-x)^0=-\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}(1-x)^k f(S')=-((1-(1-x))^n-(1-x)^0)\]and thus \[f(S)=1,\]as desired.

My solution : We use strong induction to solve the problem. The base cases are found to be true. Let for any polygon $P$ let $F(P) = x^{a(p)}(1-x)^{b(p)}$. Assume for any set $L$, with $|L| < |S|$, we have $\sum_{\substack{P \in L \\ P \ne \emptyset }}^{{}} F(P) = 1 - (1-x)^{|L|}$ Now consider $S$, if it forms a convex polygon the claim can easily be proved. Else let $\mathcal{C}$ be the convex hull of $S$. Let $|\mathcal{C}| = t$ and let the number of points inside it be $m$. So $|S| = m+t$. Let $\mathcal{P}_i $ denote the polygons which has exactly $i$ number of vertices in the convex hull $\mathcal{C}$. Note that $P_i \ne \emptyset $ for all $i$. We define $f(N_i)= \sum_{{}}^{{}} F(\mathcal{P}_i ) $ where the summation is taken over all the polygons having exactly $i$ number of vertices in the convex hull $\mathcal{C}$. Note that if $i=t = |\mathcal{C}|$, there is no polygon other than the convex hull itself containing all the points of $\mathcal{C}$ as the vertices. So $f(N_t) = F(\mathcal{C}) = x^{t}$. So $\sum_{\substack{P \in S \\ P \ne \emptyset}}^{{}}F(P) = \sum_{i=0}^{t} f(N_i ) = \sum_{i=0}^{t-1} f(N_i ) + x^t $. Note that $x^t + \sum_{i=0}^{t-1} f(N_i ) = x^t + \sum_{i=0}^{t-1}f(N_i)\left (\sum_{r=1}^{t-i} (-1)^{(r+1)} \binom{t-i}{r} \right ) $. $= x^t + \sum_{r=1}^{t}(-1)^{(r+1)}\sum_{i=0}^{t-r}\binom{t-i}{r}f(N_i)$ $= x^t + \sum_{r=1}^{t} (-1)^{r+1} \sum_{\substack{P \in S\setminus \mathcal{C}\cup T , P \ne \emptyset \\ T \subseteq \mathcal{C} \\ |T| = t-r}}^{{}} F(P)$ $ = x^t + \sum_{r=1}^{t} (-1)^{(r+1)}\binom{t}{t-r} \left ( 1- (1-x)^{t-r+m} \right )\left ( 1-x \right )^r = 1 -(1-x)^{t+m}$ $ = 1 -(1-x)^{|S|} $, and the induction is complete.

We randomly color the point by black and white, and the probability for a points to be colored by black is $p$. For each convex polygon $P$, let $\mathbb{E}[P]$ be the event that every vertices of $P$ is black and every point outside $P$ is white. These event are pairwise mutually exclusive. Hnece the LHS is the probability of the event that some $\mathbb{E}[P]$. But this surely happens, by considering the convex hull of all balck points.

We prove the statement when those $ |S| $ points form a convex $ |S| $-gon. It is easy because the convex hull of every $ P \subseteq S $ is exactly $ P $ (i.e. $ a(P) + b(P) = |S| $), so we need to evaluate: $$ \sum_{P}{x^{a(P)}(1 - x)^{b(P)}} = \sum_{P}{x^{a(P)}(1 - x)^{|S| - a(P)}} = \sum_{i=0}^{|S|}\sum_{a(P) = i}{x^{i}(1 - x)^{|S|-i}} = \sum_{i=0}^{|S|}{{|S| \choose i}x^{i}(1 - x)^{|S|-i}} = (x + (1 - x))^{|S|} = 1 $$ Let us call one point $ T $. We draw $ {|S|-1} \choose 2 $ lines for every pair of points which are not $ T $ and consider translation of $ T $ over a line $ l $ which passes through points $ A $ and $ B $ and divides plane in two halfplanes where $ H_1 $ and $ H_2 $ denote set of points in a halfplane ($ A $ and $ B $ included). Let $ f(P) \equiv x^{a(P)}(1 - x)^{b(P)} $. We can say that $ f(P) = 0$ for non-convex $P$. WLOG $ T $ goes from $ H_1 $ to $ H_2 $. $ X' $ denotes an object $ X $ after this transformation. These facts are true: $$ f(P') = f(P) \qquad \qquad \forall P \subseteq S \quad A \notin P \lor B \notin P $$because moving point $T$ over $ l $ does not affect convexity of $P$, $ a(P) $ nor $ b(P) $. $$ f(P') = f(P)(1-x) \qquad \qquad \forall P \subseteq H_1 \setminus \{ T \} \quad A, B \in P $$because moving point $T$ over $ l $ does not affect convexity of $P$ and $ b(P) $ increases by $1$ (that is point $T$). $$ f(P' \cup \{ T \} ) = f(P)x \qquad \qquad \forall P \subseteq H_1 \setminus \{ T \} \quad A, B \in P $$because $ a(P) $ increases by $1$ (that is point $T$). $$ f(P \cup \{ T \} ) = 0 \qquad \qquad \forall P \subseteq H_1 \setminus \{ T \} \quad A, B \in P $$because it is not convex. Combining last three facts and deducing similarly for $ P \subseteq H_2 \setminus \{ T \} $ we conclude that: $$ f(P') + f(P' \cup \{ T \} ) = f(P) + f(P \cup \{ T \} ) \qquad \qquad \forall P \subseteq S \setminus \{ T \} \quad A, B \in P $$ (We used that $ A, B \in P $ implies either $ P \subseteq H_1 \setminus \{ T \} $ or $ P \subseteq H_2 \setminus \{ T \} $.) Finally, $$ \sum_{P'}{x^{a(P')}(1 - x)^{b(P')}} = \sum_{P'}{f(P')} = \sum_{\substack{P' \subseteq S \\ A \notin P' \lor B \notin P'}}{f(P')} + \sum_{\substack{P' \subseteq S \setminus \{ T \} \\ A, B \in P'}}{f(P') + f(P' \cup \{ T \})} = \sum_{\substack{P \subseteq S \\ A \notin P \lor B \notin P}}{f(P)} + \sum_{\substack{P \subseteq S \setminus \{ T \} \\ A, B \in P}}{f(P) + f(P \cup \{ T \})} = \sum_{P}{f(P)} = \sum_{P}{x^{a(P)}(1 - x)^{b(P)}} $$ proving that this is a monovariat expression. We can achieve every configuration by using that transformation. Q.E.D.

Lemma If we have a set $A$ of $s$ points such that any convex polygons formed by the points not in $A$ does not contain any points in $A$ in the interior, then \[\sum_{P, A \in P}{x^{a(P)}(1 - x)^{b(P)}} = x^{|A|}.\]Proceed via induction on number of points total for this and the main problem together. Consider each distinct set $B$ of points in $S-A$ such that $A+B$ is a convex polygon. We consider the smaller set of points on and inside $A+B$, excluding the points $A$. For all convex polygons $C$ such that convex hull of $C+A$ is the same as $B+A$, $C$ is in the set. Now the sum \[\sum_{C}{x^{a(C)}(1 - x)^{b(C)}}\]by induction is $x^{|B|}$. Then readding all the other points besides $A$, we get $x^{|B|}(1-x)^{b(A+B)}$. By induction, the sum across all such $B$'s is $1$. So if we then add back $A$, we get $x^{|A|}$. Take a point $T$ on the convex hull of $S$, by induction hypothesis for lemma all the polygons with $T$ in it contribute $x$ to $\sum_{P A \in P}{x^{a(P)}(1 - x)^{b(P)}}$. And if we consider removing $T$, then the sum for the remaining points would be $1$ by inductive hypothesis so that adding back $T$ these would contribute $1-x$ to the sum ($T$ is outside all those polygons). And finally $x+(1-x) = 1$, so our sum is $1$ as desired.

Very short and nice problem. The $x$ and $1-x$ begs us to use some sort of probability construction. First assume $0<x<1$, we color the points with the following method: each point has probability that a point is red is $x$, blue is $1-x$. The expression is the probability that the convex hull of all the red points is P. Obviously summing over all P, we get the probability that the convex hull of all the red points is a convex polygon in S which is clearly true. The polynomial is 1 for $0<x<1$. Since the polynomial is 1 for an infinite number of x, and has a finite degree, It must be 1 for all x.

Lemma: Given a set $S$, and $a$ consecutive points along its convex hull, then the sum over all convex polygons containing these $a$ points is $x^a$. We will prove this with induction. The base case for the claim is when $S$ is a set with exactly $a$ points on its convex hull. In this case, there exists only one such polygon, and its value is just $x^a(1-x)^0=x^a$ as desired. Now, for the inductive step, consider an arbitrary set $S$, and $a$ points along its convex hull. Consider the next point after these $a$ points, still along the convex hull - $B$. If we sum over all convex polygons containing our $a$ points but not $B$, we will get $x^a(1-x)$, since inductive hypothesis tells us we get $x^a$ if we sum over the set $S\setminus B$, but now $B$ will add $1$ to the $b$ value of all polygons. On the other hand, if we include $B$, then inductive hypothesis from the $a+1$ case tells us that this yields a net sum of $x^{a+1}$. Combining these two cases, the sum over all convex polygons with these $a$ points is $x^a(1-x+x)=x^a$, as desired. So, the induction is complete. To finish, we will use induction on the size of $S$. The base case of $|S|=3$ is clear, so continue to $|S|=n$. In this case, consider a point $P$ on the convex hull of $S$. Using similar logic as before, the sum over all convex polygons not including $P$, by inductive hypothesis, is just $1-x$. On the other hand, the sum over all polygons with $P$ is $x$, by our Lemma, so the total sum is $1$, as desired.

There must be some method without using convex hulls

The result is obvious if $S$ consists of vertices of a convex polygon since it becomes the binomial theorem. Now imagine moving points of $S$ around; our goal is to show the LHS does not change. Let $X$ be a point and consider the $\binom{|S|-1}{2}$ lines determined by two points in $S$. As $X$ moves, there is no change unless some line is crossed. Now imagine $X$ crosses a line segment $AB$. If $P$ was a polygon that contained $A$ and $B$ as consecutive vertices, then that polygon has one more point outside it; at the same time, a new convex polygon with $a(P)+1$ vertices is created, with $b(P)$ vertices. So we need to check \[ \underbrace{x^{a(P)} (1-x)^{b(P)}}_{\text{original } P} = \underbrace{x^{a(P)} (1-x)^{b(P)+1}}_{\text{modified } P} + \underbrace{x^{a(P)+1} (1-x)^{b(P)}}_{\text{new polygon}} \]which is true. The same argument (in reverse) applies for polygons $P$ which have $A$ and $B$ as consecutive vertices in which $X$ moves in to the polygon. This handles the case where $X$ crosses a segment $AB$. If $X$ crosses the extension of ray $AB$ past $B$, we can repeat the same proof with the roles of $B$ and $X$ swapped. Hence, we finally conclude that moving $X$ does not change the LHS. Since we may move any point in the set arbitrarily, this shows the LHS depends only on $|S|$ and hence must equal $1$.

I guess this is a repeated solution..but can't help! Let us just consider $x \in [0,1]$, because that is enough to imply the problem for all reals. We colour every vertex of $S$ red or blue with probability $x$ and $1-x$ respectively. Then, \[{x^{a(P)}(1 - x)^{b(P)}} \]is just the probability that every vertex of $P$ is red and all points outside $P$ are blue. When we take a sum over all convex polygons, the resulting sum of the probabilities is just the probability of that there is at least one such convex $P$ having all vertices red and all points outside it blue. But such a $P$ always exists (the convex hull of all red points, which may be empty but that is allowed). We are done.

Color each point red with probability $x$ and blue with probability $1-x$. Call a polygon $P$ good if all its vertices are red and all the vertices outside it are blue. Then \[ \Pr[\text{$P$ good}] = x^{a(P)}(1-x)^{b(P)}. \]So \[ \mathbb{E}[\text{number of good polygons}] = \sum_{P} 1\cdot \Pr[\text{$P$ good}] = \sum_{P} x^{a(P)}(1-x)^{b(P)}. \]where the expectation is taken over all possible colorings. But note that in a fixed coloring, there is exactly 1 good polygon: the convex hull of the red points. So the expectation over all polygons is also 1, finishing.

Solved with GeronimoStilton and awang11: Because polynomials are happy, assume $x \in [0,1]$. Color every point red randomly with probability $x$. Then $\sum_{P} x^{a(P)}(1-x)^{b(P)}$ is the probability that some $P$ is the convex hull of the red points. This is $1$.

Let $x$ be the probability that we color a certain point white, and then $1 - x$ probability of black. The left hand basically just computes the probability of a specific polygon being a white convex hull. Exactly one exists, so we are done.

Let $a=x$ and $b=1-x$, notice that for each round subset $A$ let $I(A)$ be the number of points in its interior then $$P(x)=\sum_{\text{A round}}a^{|A|}b^{r(A)}(a+b)^{I(A)}\hspace{20pt}(1)$$Notice that if we expand the right hand side of $(1)$, we get a homogeneous polynomial with degree $|M|$. Let $|M|=k$ we show that the coefficient of the term $a^ib^{k-i}$ has coefficient $\binom{k}{i}$. Indeed, we create a bijection between each such terms and $i$-point sets in $M$. Consider every $i$-point set in $M$, let $A$ be its convex hull then it will be a round subset, then it will correspond to the term $a^{|A|}b^{r(A)}a^{i-|A|}b^{I(A)-i+|A|}$. It is easy to see that this is a bijection. Therefore, $$P(x)=\sum_{i=1}^n\binom{n}{i}a^ib^{n-i}=(a+b)^n=1$$as desired.

Feels somewhat contrived, as the summation idea kind of implies the solution. It suffices to prove the identity for $0<x<1$. Define a set $A \subset S$ of points as follows: any given point in $S$ has a probability $x$ of being in $A$. Then, the sum on the left is equal to the expected number of polygons $P$ for which no points in $A$ are excluded from $P$, and all points on the boundary of $P$ lie in $A$. However, this quantity is precisely 1 because the only polygon that satisfies these conditions is the convex hull of $A$, which proves the identity.

First suppose $x \in (0,1)$. Randomly color the points in $S$ red with probability $x$ and blue with probability $1-x$. I claim both sides count the expected number of convex polygons with entirely red vertices such that every red vertex lies within that polygon. Summing over polygons (by linearity of expectation) yields the summation on the LHS, since for a polygon $P$ to satisfy the conditions described, we need $a(P)$ specific vertices to be red and $b(P)$ specific vertices to be blue. On the other hand, there will always be exactly one polygon $P$ that works: the convex hull of the red points. This is because if some part of this convex hull is not in $P$, then a red point lies outside $P$ (a vertex of said convex hull), and if $P$ contains the convex hull and some other vertex, that vertex must be blue. To finish, note that for any fixed $S$, the equation is a polynomial identity. Since both sides agree for infinitely many $x$, they must agree for all $x \in \mathbb{R}$, so we're done. $\blacksquare$ Remark: Feels like the condition about $P$ being convex is a bit of a giveaway regarding the involvement of a convex hull?

We choose to select or not select each point independently and at random with probability $x$. Then, the probability that all the vertices of $P$ is selected is $x^{a(P)}$, and the probability that no point outside of $P$ is selected is $(1-x)^{b(P)}$. Thus, the probability that $P$ is the convex hull of the selected points is \[x^{a(P)}(1 - x)^{b(P)}\]Each set of points has exactly one convex hull, so we are done.

We will prove that the LHS is equal to $1$ for any $x$ such that $0<x<1$, then the result follows automatically because the LHS is a polynomial. Consider colorings of the points in $S$, with each point being red with probability $x$ and blue with probability $1-x$, and the colors being independent. Call a polygon good for a coloring if all of its vertices are red and all of the points outside it are blue. Claim. The convex hull of the red points, $P$, is the only good polygon. Proof. Obviously $P$ is good. Now consider a polygon $Q\ne P$ that has a vertex set that is not a superset of the vertex set of $P$. Then choose a vertex $v$ that $Q$ left out but is a vertex in $P$. The only way to not have $v$ be on the outside of $Q$ is for $Q$ to take points that are outside the convex hull, which are not red, so $Q$ is bad. Now consider a polygon $Q\ne P$ that has a vertex set that is a superset of the vertex set of $P$. Then $Q$ contains points outside the convex hull, which are not red, so $Q$ is bad. Now simply note that a coloring makes a specific polygon good with probability $x^{a(P)}(1-x)^{b(P)}$. QED.

Silly funny goofy problem Suppose all the points are initially grey, and we color some of the points in $S$ with probability $p$. First, notice that there is always a unique convex polygon $Q$ with all red vertices and with no red points outside of $Q$, just take $Q$ to be the convex hull of the red points. But, given any polygon $P$, the probability that all the vertices of $P$ are red and all the points in $S$ outside of $P$ are grey is $p^{a(P)}(1-p)^{b(P)}$. But since such a polygon must exist, it must be the case that \[ \sum_{P} p^{a(P)}(1-p)^{b(P)} = 1 \]

This might be one of my favourite double counting problems. For each convex polygon $P$ with vertex set $V(P)$ contained in $S$, define $I(P)$ to be the set of points lying inside $P$, $E(P)$ the set of points lying outside $P$. Also denote by $c(P)$ the cardinality of $I(P)$. We call the wanted value $A$. Set $y=1-x$. Now we begin our magical calculation. \begin{align*} A&=\sum_{P}x^{a(P)}y^{b(P)} \\ &=\sum_{P}x^{a(P)}y^{b(P)}(x+y)^{c(P)} \\ &=\sum_{P}x^a(P)y^{b(P)}\sum_{J\subset I(P)}x^{|J|}y^{c(P)-|J|} \\ &=\sum_{P}\sum_{J\subset I(P)}x^{a(P)+|J|}y^{b(P)+c(P)-|J|} \\ &=\sum_{K\subset S}x^{|K|}y^{|S|-|K|} \\ &=(x+y)^{|S|} \\ &=1 \end{align*} Then we are done!

One of the problems of all time. Let $c(P)$ be the number of points inside $P$, such that $a(P) + b(P) + c(P) = |S|$. Then, it remains to show that \[ \sum_{P}{x^{a(P)}(1 - x)^{b(P)}(x + (1-x))^{c(P)}} = 1 \] Consider $2$-colorings $(R, B)$ of $S$ where vertices in $P$ are painted red, vertice outside $P$ are blue, and vertices inside can be painted either color, where $R$ and $B$ are the number of each color painting respectively. The LHS is equivalent to summing $x^R(1-x)^B$ over all two colorings. Then, there exists a bijection between the possible values of $R$ and $\mathcal{P}(S)$ since for each $s \in \mathcal{P}(S)$, it corresponds to letting $P$ be the convex hull of $s$ and coloring the interior points accordingly. As such, the sum is just \[ \sum_{P \in \mathcal{P}(S)} x^{|P|}(1-x)^{|S \setminus P|} = ((x) + (1-x))^{|S|} = 1. \]

Assume that $x \in (0, 1)$ since the LHS is a polynomial in $x$ (if the identity holds for infinitely many reals, it holds for all reals). Color the points in $S$ black with probability $x$ and white with probability $1 - x$. Then the LHS sums the probability that all of the vertices of $P$ are black while all of points outside $P$ are white, across all convex polygons $P$. But for every coloring of the points, there exists exactly one such polygon; the convex hull of the black points (all other convex polygons $P'$ either have white points as vertices or black points lying outside $P'$). Therefore the LHS indeed equals $1$ for all $x \in (0, 1)$, so it must be $1$ for all reals and we are done.

Fortunately the polynomial given in the problem is not a series, so convergence issues are off our back. In particular, it suffices to prove the problem for $x \in (0, 1)$. That said, consider the following random process. Color each point in $S$ red with probability $x$ and blue with probability $1-x$. The quantity in the LHS of the desired equality is the probability that the convex hull of the red points is a convex polygon. But clearly this probability is $1$ as no three points in $S$ are collinear, and we conclude.

Pick each point with probability $x$, and pick the convex hull of these points. We claim that the summand is exactly the probability that the polygon in question is picked. First of all, we must pick all of the vertices of the polygon itself, which is $x^{a(P)}$. Then, we must not pick any vertices in the exterior of the polygon, which happens with probability $(1-x)^{b(P)}$. We do not care about the status of the vertices inside the polygon, since we are taking the convex hull. Thus, the sum of the probabilities is 1, which solves the problem for $0\leq x\leq 1$. To finish, note that the summation is a polynomial in $x$, and it matches the constant polynomial $1$ at infinitely many values, so by identity theorem it is always 1.

Choose a subset of the points randomly by including each point with probability $x$. A given convex polygon $P$ is their convex hull if and only if all vertices of $P$ are chosen and all points outside $P$ are not chosen, with probability $x^{a(P)}(1-x)^{b(P)}$. These probabilities must sum to $1$, so we are done.

storage

We colour the points with red with probability $x$ and blue with $1-x$. Let a convex polygon $P$ with red vertices and all vertices outside to be blue, be called nice. Then: in LHS, each term is probability of having a convex polygon $P$ to be nice and thus, is the expected value of number of polygons to be nice. But for any random configuration, there is only one valid nice polygon $P$ which is the convex hull of the red points.