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As an outcome of this question

what does the Inclusion–exclusion principle means in disjoint? is {1,4}$\cap${1,2}=$\emptyset$?

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gbox
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  • The intersection is ${1}$. – André Nicolas Jun 19 '14 at 16:46
  • so what is the sum of $|A \cup B|$? A={1,4},{2,4},{3,4} B={1,2},{2,3},{1,3}? |A|+|B|-$|A \cap B|$= 3+3-6? – gbox Jun 19 '14 at 16:48
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    Clearly, if A={{1,4},{2,4},{3,4}} and B={{1,2},{2,3},{1,3}} $|A|=3$ and $|B|=3$. Due to the fact that A and B are disjoint (i.e. $A \cap B = \emptyset$), then $|A \cup B|= 3+3=6$. – Mauro ALLEGRANZA Jun 19 '14 at 16:53
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    $A$ and $B$ are probably intended to be collections (sets) of sets, though you left out the outer braces. The union has $6$ elements. But that does not get at Inclusion/Exclusion. – André Nicolas Jun 19 '14 at 16:54
  • see here http://math.stackexchange.com/questions/839726/combinatorial-proof-of-n-choose-k-n-1-choose-k-1n-1-choose-k/839748#comment1732893_839732 – gbox Jun 19 '14 at 16:58

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The inclusion-exclusion principle tells you $|\{1,4\}\cup\{1,2\}|=|\{1,4\}|+|\{1,2\}|-|\{1,4\}\cap\{1,2\}|$

The inclusion-exlusion principle is not usually very useful when looking at explicit sets, it is more useful for proving combinatorial identities via counting two ways.

Asinomás
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  • see here http://math.stackexchange.com/questions/839726/combinatorial-proof-of-n-choose-k-n-1-choose-k-1n-1-choose-k/839748#comment1732893_839732 – gbox Jun 19 '14 at 16:59
  • The inclusion exlclusion principle tells you he size of set A union B is the same as the size of A plus the size of B minus the size of the intersection, it doesn't matter what size A,B or A intersected with B have, this always works. – Asinomás Jun 19 '14 at 17:04