Let $A_{1},\ A_{2}$, , $A_{n}$ be $n$ sets. Then $|A_{1}\cup A_{2}\cup … \cup A_{n}|=
\sum_{i}|A_{i}|-\sum_{i<j}|A_{i}\cap A_{j}|+ \cdots + (-1)^{n-1}|A_{1}\cap A_{2}\cap\ \cap A_{n}|.
$Proof. Suppose that $x\in \bigcup_{1 \le h \le n} A_h $ is in
exactly $m$ of the sets $A_{1}, A_{2}$, . . . $, A_{n}$. How many of the sets of the form $A_{i_{1}}\cap A_{i_{2}}\cap\cdots\cap A_{i_{k}}$ with $i_1 < i_2 < \cdots < i_k$ is $x$ in? The answer is $\left(\begin{array}{l}
m\\
k
\end{array}\right)$ because it is just the number of ways of choosing $k$ of the sets which contain $x$.$\color{red}{\blacklozenge}$ This means that on the right hand side of the formula, $x$ will be counted
$
\left(\begin{array}{l}
m\\
1
\end{array}\right)\ -\ \left(\begin{array}{l}
m\\
2
\end{array}\right)\ +\cdots +(-1)^{m-1}\ \left(\begin{array}{l}
m\\
m
\end{array}\right)
$ times.
I espy why the IEP and the last line (after the red lozenge) are true for $m \le 3$ sets, thanks to pictures like this one on page 2 of 7. Yet I don't perceive it when $m \ge 4$ (which I don't think can be visualised) ? Whence does this expression originate?
Best Answer
You are right saying that no similar picture can be made with discs. However a set can be given any shape. Here's an example with four sets.