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I know there are answers to this specific question, but I'm trying to solve it through a specific line of thought which was used in a textbook on modern algebra for the case with 2 sets. It goes like this:

For two finite sets, let $A'$ and $B'$ be the parts of $A$ and $B$ which have elements that are NOT in $A \cap B$. The number of elements of the union of the two sets, denoted by $n ( A \cup B )$, can be found by:

$ n(A \cup B) = n(A') + n(B') + n(A \cap B)$

But we have: $n(A') = n(A) - n(A \cap B)$ and $n(B') = n(B) - n(A \cap B)$. Those two relations plugged back in the first expression give us the correct answer:

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Now to the case with 3 sets $A$, $B$ and $C$. I followed the same reasoning, taking the parts $A'$, $B'$ and $C'$ that are not in any of the intersections and added them up:

$ n(A \cup B \cup C) = n(A') + n(B') + n(C') + n(A \cap B) + n(A \cap C) + n(B \cap C) + n(A \cap B \cap C)$

Where we have:

$n(A') = n(A) - n(A \cap B) - n(A \cap C) - n(A \cap B \cap C)$

$n(B') = n(B) - n(A \cap B) - n(B \cap C) - n(A \cap B \cap C)$

$n(C') = n(C) - n(A \cap C) - n(B \cap C) - n(A \cap B \cap C)$

Plugging it back those relations I find an incorrect answer with a factor of $-2$ multiplying $n (A \cap B \cap C)$, that is:

$n(A \cup B \cup C ) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) - 2 n (A \cap B \cap C)$

Can someone point out the mistake in this reasoning? I'd appreciate very much.

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  • $\begingroup$ Note that $A \cap B \cap C \subset A \cap C$ and so on. $\endgroup$ – AugSB Feb 21 '17 at 17:13
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When you split up the sum into $n(A') + n (B')+.. + n(A\cap B \cap C)$, you are over counting $A\cap B \cap C$ three times since it is already in the pairwise intersections.

You make similar mistakes further on as well. They all stem from this fact though.

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