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Let $A,B,$ and $C$ be sets such that

$A\subset C, B\subset C$, with $|C|=n$, $|A|=x$, $|B|=y$, and $|A\cap B|=z$.

What is $|C\setminus(A\cup B)|$?

ervx
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starsaber99
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    So exactly which part of your question is about notation? – Kenny Lau Aug 08 '18 at 13:22
  • I guess notation wasn't the right word, but I'm a little confused about what "C" means and how the fact that A and B are subsets of C affect the question? I mean, I know that all the elements of A and B are within the set C, so I guess it's really just "C" confusing me. – starsaber99 Aug 08 '18 at 13:23
  • The fact that they're subsets of $C$ affects the question in that it allows the question to be answered. Without this premise, you wouldn't have enough information to determine $|C\setminus(A\cup B)|$. – joriki Aug 08 '18 at 13:25
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    It's the relative complement: $C\setminus A={x\in C|x\notin A}$ – saulspatz Aug 08 '18 at 13:27
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    So the relative complement between C and $(A\cup B)$ is basically all the elements that are in C but not in $(A\cup B)$? – starsaber99 Aug 08 '18 at 13:31
  • @starsaber99: Yes. – joriki Aug 08 '18 at 13:35
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    So with regards to the variables given, it would be written as: $n-(x+y-z)$? – starsaber99 Aug 08 '18 at 13:38

1 Answers1

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Using Inclusion-Exclusion:

$$ |A\cup B|=|A|+|B|-|A\cap B|=x+y-z. $$

Now, $C\setminus (A\cup B)$ is the set containing all things in $C$ that are not in the set $A\cup B$. Since $C$ contains $n$ elements, and $A\cup B$ contains $x+y-z$ elements, it follows that

$$ |C\setminus (A\cup B)|=n-(x+y-z)=n+z-x-y. $$

ervx
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