An each summa5on is over all i pairs ij with ij

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Unformatted text preview: sum rule. Why? … because we would be over coun5ng the number of possible outcomes. •  Instead we have to count the number of possible outcomes of e1 and e2 minus the number of possible outcomes in common to both; i.e., the number of ways to do both tasks •  If again we think of them as sets, we have |A1 ∪ A2| =|A1| + |A2| - |A1∩ A2| CSCE 235 Combinatorics 7 PIE (2) •  More generally, we have the following •  Lemma: Let A, B, be subsets of a finite set U. Then 1.  2.  3.  4.  5.  |A∪B| = |A| + |B| - |A∩B| |A ∩ B| ≤ min {|A|, |B|} |A\B| = |A| - |A∩B| ≥ |A|- |B| |A| = |U| - |A| |A⊕B| =|A∪B|- |A∩B|= |A|+|B|- 2|A∩B|= |A\B|+ |B \A| 6.  |A × B| = |A|×|B| CSCE 235 Combinatorics 8 PIE: Theorem •...
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This note was uploaded on 10/31/2013 for the course CSCE 235 taught by Professor Staff during the Spring '08 term at UNL.

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