Q:

Why is the number of elements in a union of 3 sets that? 

In my maths notebook, it says that if $A$, $B$, and $C$ are finite sets then:
$$ 
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
$$
However I don't know how this works properly.  It does seem to work based on the previous identities on union of sets (i.e. $n(A ∪ B) = n(A) + n(B) – n(A ∩ B)$).  However when I try to visualize it in Venn Diagram form, it just doesn't seem to work.  I'm guessing this is because of my misconception between area and count of elements, but I still don't know why it does work.  Any sort of help that helps me understand how to think of this equation will be greatly appreciated. 

A:

Here is a $3$ circle Venn diagram.  Starting from top to bottom, left to right, I will label the colors:

$|A ∩ B' ∩ C'|$ ($A$ and nothing else): blue
$|A ∩ B ∩ C'|$: dark green
$|A' ∩ B ∩ C'|$ ($B$ and nothing else): green
$|A ∩ B' ∩ C|$: red
$|A ∩ B ∩ C|$: brown
$|A' ∩ B ∩ C|$: orange
$|A' ∩ B' ∩ C|$ ($C$ and nothing else): pink

For the union of all of these sets, we can only have exactly one of each region. 
Notice that:
$$
\begin{align}
|A| &= \text{blue} + \text{dark green} + \text{red} + \text{brown}\\
|B| &= \text{green} + \text{dark green} + \text{orange} + \text{brown} \\
|C| &= \text{pink} + \text{red} + \text{orange} + \text{brown} \\
|A| + |B| + |C| &= \text{blue} + \text{green} + \text{pink} + 2(\text{dark green} + \text{orange} + \text{red}) + 3\text{brown}
\end{align}
$$
We need to subtract the overlaps of two circles. 
$$
\begin{align}
|A∩B| &= \text{dark green} + \text{brown}\\
|B∩C| &= \text{orange} + \text{brown} \\
|A∩C| &= \text{red} + \text{brown} \\
|A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| &= \text{blue} + \text{green} + \text{pink} + \text{dark green} + \text{orange} + \text{red}
\end{align}
$$
We need to add that last brown region back. 

A:

Use Venn diagram.  First sum $n(A)$, $ n(B)$ and $n(C)$ as you did.  Then their intersections in pairs are counted twice, so substract each of them once, i.e. substract $n(A\cap B), n(A\cap C)$ and $ n(B\cap C)$ .  Initially you added the triple intersection three times, and later substracted it three times.  So you have to add it again, $n(A\cap B\cap C)$.

