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 $S(i)$}}} $$