Hint:
For $A\cup B \cup C$ you get $(|A\cap B| - |A\cap B\cap C|) + (|A\cap C| - |A\cap B\cap C|) + (|B\cap C|- |A\cap B \cap C|) = |A\cap B| + |A\cap C| + |B\cap C| - 3|A\cap B\cap C|$
For $A\cup B \cup C \cup D$ you get $\sum_{\{X,Y\}\subset\{A,B,C,D\}\text{ there are six of these}} (|X\cap Y| - (\sum_{Z\in \{A,B,C,D\}\setminus\{X,Y\}\text{ there are two of these}} |X\cap Y\cap Z|)+(|A\cap B\cap C\cap D|_{\text{ there is one of these}}) = $ the six $|A\cap B|, |A\cap C|$ etc... + each of the $|A\cap B\ \cap C|$ etc appear each in exact $3$ of the summands (that is, $A\cap B \cap C$ will be subtracted from each of the three: $A\cap B$, $A\cap C$ and $B \cap C$. or $3$ times total).. + $A\cap B\cap C\cap D$ appearing in each of the $6$ times.
So.... So that is where the coefficients come from.
So for $A\cup B \cup C \cup D \cup E$ you will have:
$10= {5 \choose 2}$ combinations of $X \cap Y$.
Each will substract some $X\cap Y\cap Z$. For each of the $X\cap Y$ there will be $3$ of these. (Example: for $A \cap B$ we will remove $A\cap B \cap C$ and $A\cap B \cap D$ and $A\cap B \cap E$; that is $3 = 5-2$ of them.) Of the ${5\choose 3} = 10$ combinations of $X\cap Y\cap Z$, $10*3= 30$ will be subtracted so each of them will be subtracted $\frac {30}{10} = 3$ times.
For each of the $X \cap Y$ you will subtract some of the $X\cap Y\cap Z\cap W$. There will but ${(5-2)\choose 2}={3\choose 2}= 3$ of these. (That is. For $A\cap B$ there will be $A\cap B \cap C\cap D$ or $A\cap B \cap C\cap E$ or $A\cap B \cap D\cap E$). So we will subtract $10*3=30$ of these 4-sects. There are ${5\choose 4} = 5$ of these 4-sects so each will be subtracted $\frac {30}4 = 6$ times.
And the $A\cap B \cap C \cap D \cap E$ will be subtracted $10$ times.
In general: TO find the elements in exactly $2$ of $n$ sets.
You will have ${n \choose 2}$ $A\cap B$ to consider.
For each of those you will have $n-2$ $A\cap B \cap C$ to subtract. That is ${n \choose 2}*(n-2)$ total so each of the $n \choose 3$ of the $A\cap B \cap C$ will each be subtracted $\frac {{n \choose 2}*(n-2)}{n\choose 3} = \frac {n!(n-3)!3!}{(n-2)!2!n!}(n-2)=3$
For each of the ${n \choose 2}$ $A\cap B$ that we consider there are ${n - 2\choose k}$ $A \cap B \cap .... \cap K$. That is ${n \choose 2}*{n - 2\choose k}$ so for each of the $n\choose k$ of the $A\cap B\cap ... K$ each will be subtracted /added $\frac {{n \choose 2}*{n - 2\choose k}}{n \choose k}={n-k \choose 2}$.
Basically. The coefficients for the elements of $n$ sets that occur in exactly $m$ sets will be ${n-k \choose m}$.
