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Given $$n(A) = 2600 ,n(B)=1200, n(C)=2500$$
and $$n(A \cap C)=1000, n(A \cap B)=400, n(B \cap C)=300, n(A \cap B \cap C)=200$$ Determine $$n(B \cap \bar{A} \cap \bar{C})$$

It's been ages since I've done set theory, and despite knowing DeMorgan's Laws as well as basic two-set laws, I can't figure out how to solve this purely algebraically without resorting to a long winded implementation of $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$ It's pretty easy to see using a Venn diagram or breaking it down using this equality that the solution is $700$, but I feel like I'm missing a much simpler logic or set-theory law.

My question is, is there a set-theory law or course of action to make this question relatively simple to answer algebraically?

Any help is greatly appreciated!

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  • $\begingroup$ Notationally, is $n$ cardinality and $\overline{A}$ the complement of $A$? $\endgroup$ – Moya Mar 10 '15 at 1:41
  • $\begingroup$ @Moya Yes, is different convention typically implemented here? If so I can edit the question. $\endgroup$ – HavelTheGreat Mar 10 '15 at 1:43
  • $\begingroup$ It depends on the context. I was taught that the cardinality is $|A|$ and the complement is $A^c$ or $X\setminus A$, for $X$ the larger, ambient set universe you're in. I understood you so you're fine. $\endgroup$ – Moya Mar 10 '15 at 1:45
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    $\begingroup$ I get 1400, not 800, but I might have been careless. In any event, the Venn Diagram is (in my opinion) definitely the best and quickest way to solve this problem. $\endgroup$ – WillO Mar 10 '15 at 1:55
  • $\begingroup$ @WillO Ahh my apologies, mis-wrote my question. That is the correct answer for what I had written though. No simple set-theory trick eh? :( $\endgroup$ – HavelTheGreat Mar 10 '15 at 1:57
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You are looking for everything that is in B, not in A, and not in C. So take whats in B, subtract out what B and C have in common, then subtract out what B and A have in common, then add back what they all have in common. Its the standard excluded middle formula.

1200 - 300 - 400 + 200 = 700

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  • $\begingroup$ Cheers, thanks a ton! $\endgroup$ – HavelTheGreat Mar 10 '15 at 2:18

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