How do I prove that for any finite subsets A and B exists one set R? How do I prove that for any finite subsets A and B exists one set R
$\left | A\cup B \right |=\left | A \right |+\left | B \right | -\left | A\cap B \right |$
Deduce from this an adequate formula for $\left | A\cup B\cup C \right |$. (With $\left | M \right |$ is the number of elements of M determined). 
My effort: I don't know where to start.
 A: Say you have a collection of toys of different shapes and colours. Let $A$ be the subset of toys that are cars, and $B$ be the subset consisting of red toys. Then $A\cup B$ is the set of toys that are either red or cars (or red cars), while $A\cap B$ is the set of red cars. Now can you make more sense of the formula
$$
|A\cup B|=|A|+|B|-|A\cap B|?
$$It says that if you want to count the number of toys that are red or cars (or both), you count the number of cars and you count the number of red toys separately. However, all the red cars have been counted twice (both as part of $A$ and as part of $B$), so the result you get is exactly $|A\cap B|$ too large. Therefore we subtract it once.
Now, let $C$ be the set of toys that use batteries. It gets a bit messier, but the idea is exactly the same. Start with $|A|+|B|+|C|$, and then look at which types have been counted twice and thrice. Subtract them, but note that you have now subtracted some toys too many times, and they haven't been counted at all. Add them back, and you're done.
A: For the second part:
We know for any finite sets $X$ and $Y$ that $$|X\cup Y|=|X|+|Y|-|X\cap Y|. \qquad (*)$$
So, from $(*)$ with $X=A$ and $Y=B\cup C$, we have
$$|A\cup B\cup C|=|A\cup (B\cup C)|=|A|+|B\cup C|-|A\cap(B\cup C)|.\qquad (1)$$
From $(*)$ with $X=B$ and $Y=C$ we also have $$|B\cup C|=|B|+|C|-|B\cap C|.\qquad (2)$$ Also, since
$$A\cap(B\cup C)=(A\cap B)\cup (A\cap C)$$ you have from $(*)$ with $X=A\cap B$ and $Y=A\cap C$ that $$|A\cap(B\cup C)|=|A\cap B|+|A\cap C|-|(A\cap B)\cap (A\cap C)|.$$
But $(A\cap B)\cap (A\cap C)=A\cap B\cap C$, so 
$$|A\cap(B\cup C)|=|A\cap B|+|A\cap C|-|A\cap B\cap C|.\qquad (3)$$
Substituting $(2)$ and $(3)$ into $(1)$ gives the formula:
$$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|.$$
