Demonstrate another way to implement the Inclusion–exclusion principle?
Let's derive a generalization of the Inclusion-Exclusion Principle based on some combinatoric identities.
Cancellation Lemma
Suppose $n$ and $k$ are non-negative integers. Then $$ \sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k} =\left\{\begin{array}{} 1&\text{if }n=k\\ 0&\text{otherwise} \end{array}\right.\tag{1} $$
Note that if $n\lt k$, then $\binom{n}{j}\binom{j}{k}=0$ for all $j$, so we can assume that $n\ge k$.
Proof 1: Using $\displaystyle\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$, we get $$ \begin{align} \sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k} &=\sum_{j=k}^n(-1)^{j-k}\binom{n}{k}\binom{n-k}{j-k}\\ &=\binom{n}{k}(1-1)^{n-k}\tag{2} \end{align} $$ which is $1$ if $n=k$ and $0$ otherwise. $\quad\square$
Proof 2: Using Vandermonde's Identity, we get $$ \begin{align} \sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k} &=\sum_{j=k}^n(-1)^{j-k}\binom{n}{n-j}\binom{j}{j-k}\\ &=\sum_{j=k}^n\binom{n}{n-j}\binom{-k-1}{j-k}\\ &=\binom{n-k-1}{n-k}\tag{3} \end{align} $$ which is $1$ if $n=k$ and $0$ otherwise. $\quad\square$
Proof 3: Consider the generating function for $(1)$ as a function of $k$: $$ \begin{align} \sum_{k=0}^n\sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k}x^k &=\sum_{j=0}^n\sum_{k=0}^j(-1)^{j-k}\binom{n}{j}\binom{j}{k}x^k\\ &=\sum_{j=0}^n(-1)^j\binom{n}{j}(1-x)^j\\ &=(1-(1-x))^n\\[12pt] &=x^n\tag{4} \end{align} $$ Equating the coefficients of $x^k$ in $(4)$ proves the result. $\quad\square$
Theorem (Generalized Inclusion-Exclusion Principle)
Let $\{S(i)\}_{i=1}^m$ be a finite collection of sets from a finite universe.
Let $N(j)$ be the sum of the sizes of all intersections of $j$ of the $S(i)$: $$ N(j)=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|\tag{5} $$ Thus, $N(0)$ is the size of the universe.
Then, the number of elements in exactly $k$ of the $S(i)$ is $$ \sum_{j=0}^m(-1)^{j-k}\binom{j}{k}N(j)\tag{6} $$
Proof:
An element that is in $n$ of the $S(i)$ is counted $\binom{n}{j}$ times in $N(j)$. That is, there are $\binom{n}{j}$ ways to choose the $j$ sets in the intersection from the $n$ sets that contain the element. The Cancellation Lemma says that this element is counted once in $(6)$ if $n = k$ and is cancelled out otherwise. $\quad\square$
Using the identity $$ \sum_{k=0}^n(-1)^k\binom{j}{k}=(-1)^n\binom{j-1}{n}\tag{7} $$ where $\binom{-1}{n}=$$(-1)^n\binom{n}{n}$$=(-1)^n$$[n\ge0]$, we get the following
Corollary 1:
The number of elements in at most $k$ of the $S(i)$ is $$ \sum_{j=0}^m(-1)^{j-k}\binom{j-1}{k}N(j)\tag{8} $$
Using the identity $$ \sum_{k=n}^j(-1)^k\binom{j}{k}=(-1)^n\binom{j-1}{j-n}\tag{9} $$ again where $\binom{-1}{n}=(-1)^n\binom{n}{n}=(-1)^n[n\ge0]$, we get the following
Corollary 2:
The number of elements in at least $k$ of the $S(i)$ is $$ \sum_{j=k}^m(-1)^{j-k}\binom{j-1}{j-k}N(j)\tag{10} $$
The Inclusion-Exclusion Principle
The usual Inclusion-Exclusion Principle is the case $k=1$ of Corollary 2.
However, the usual Inclusion-Exclusion Principle can also be derived by subtracting the number of objects that are in none of the $S(i)$ from the size of the universe. Using $(6)$ yields $$ \underbrace{\ \ \ N(0)\ \ \ \vphantom{\sum_j\binom{j}{0}}}_{\substack{\text{size of the}\\\text{universe}}}-\underbrace{\sum_{j=0}^m(-1)^{j}\binom{j}{0}N(j)}_{\substack{\text{number of elements}\\\text{in none of the $S(i)$}}}=\underbrace{\sum_{j=1}^m(-1)^{j-1}N(j)\vphantom{\binom{j}{0}}}_{\substack{\text{number of elements in}\\\text{at least one of the $S(i)$}}} $$
I'm not sure whether this is what you're getting at, but you can use an index set $S$ to represent which intersection is being taken. This is similar, but not identical to, your binary representation. Then the sign of a particular term is $(-1)^{\lvert S\rvert-1}$. If $[1,n]$ denotes the set $\{1,2,\ldots,n\}$, then the Principle of Inclusion-Exclusion can be written $$ \left\lvert\bigcup_{i\in[1,n]}A_i\right\rvert+{\sum_{S\subseteq[1,n]}}'(-1)^{\lvert S\rvert}\left\lvert\bigcap_{i\in S}A_i\right\rvert=0. $$ I use the prime after the summation symbol to indicate that the sum runs over all subsets of $[1,n]$ except for the empty set.
If we adopt the reasonable convention that the empty intersection equals the universal set, then we don't have to exclude the empty set from the sum, and we can write $$ \left\lvert\left(\bigcup_{i\in[1,n]}A_i\right)'\right\rvert=\sum_{S\subseteq[1,n]}(-1)^{\lvert S\rvert}\left\lvert\bigcap_{i\in S}A_i\right\rvert=0. $$