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I'm self learning math (and English as well, plz forgive my bad English). While reseaching I found this rule:

If A, B and C are three 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 )

I cannot find any explaination about this rule. Could you help to prove this equation (maybe explain by Venn diagram)? Another question, why the last element is + n( A ∩ B ∩ C ) but not - n( A ∩ B ∩ C ) (since in the case of two sets, we have n( A ∪ B ) = n(A) + n(B) - n( A ∩ B ))

Mauro ALLEGRANZA
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SoT
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    See Inclusion–exclusion principle – Mauro ALLEGRANZA Aug 29 '22 at 12:42
  • If you count the elements of A and add those of B you will count twice the common elements. This explain the two-sets case. Similar for the three-sets case... – Mauro ALLEGRANZA Aug 29 '22 at 12:43
  • It can help to build intuition if you look at a venn diagram where each region is individually labeled. For example this one (not the greatest image, but first I found). Using the labeling on the image, note that $n(S) = n(1)+n(2)+n(4)+n(5)$, that $n(P)=n(2)+n(3)+n(5)+n(6)$, that $n(S\cup P\cup M) = n(1)+n(2)+\dots+n(7)$, and so on... You will find that if we were to just add $n(S)$ and $n(P)$ we will have added $n(2)$ and $n(5)$ too many times since they were both a part of $S$ and $P$. The formula you cite cancels correctly – JMoravitz Aug 29 '22 at 12:50
  • Extending to many sets, you will find that you will alternate between subtracting and adding (and then subtracting again and then adding again etc...) the pairs, triples, quadruples etc... of intersections of sets. This can all be formally proven by induction. Recognize that $(A\cup B\cup C) = (A\cup B)\cup C$, and that $(A_1\cup A_2\cup \dots \cup A_n\cup A_{n+1}) = (A_1\cup A_2\cup\dots\cup A_n)\cup A_{n+1}$ – JMoravitz Aug 29 '22 at 12:53
  • Thank you both of you, now I understand – SoT Aug 29 '22 at 13:05
  • See this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Aug 29 '22 at 14:56

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