19-Inclusion-Exclusion - EECS 210 Discrete Structures David...

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Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 1 EECS 210 Discrete Structures David O. Johnson Fall 2016
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Reminders Homework 5 due: Thursday, November 10 at the beginning of your lecture Connect 5 due: 11:59 PM, Thursday, November 17 Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 2
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Any Questions? Inclusion-Exclusion 3 David O. Johnson EECS 210 (Fall 2016)
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Inclusion-Exclusion (Section 8.5) Inclusion-Exclusion Number of Elements in 2 Finite Sets Number of Elements in 3 Finite Sets The Principle of Inclusion-Exclusion Number of Elements in n Finite Sets Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 4
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In Section 2.2, we developed the following formula for the number of elements in the union of two finite sets: The formula is includes the sum of the number elements in the sets and then excludes the number of elements in the intersection of the sets. We will generalize this formula to finite sets of any size. Inclusion-Exclusion Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 5
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Number of Elements in 2 Finite Sets Example : In a discrete mathematics class every student is a major in computer science or mathematics or both. The number of students having computer science as a major (possibly along with mathematics) is 25; the number of students having mathematics as a major (possibly along with computer science) is 13; and the number of students majoring in both computer science and mathematics is 8. How many students are in the class? Solution : | A B | = | A | + | B | −| A B | = 25 + 13 − 8 = 30 Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 6
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Number of Elements in 3 Finite Sets Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 7
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Number of Elements in 3 Finite Sets Example : A total of 1232 students have taken a course in Spanish 879 have taken a course in French 114 have taken a course in Russian 103 have taken courses in both Spanish and French 23 have taken courses in both Spanish and Russian 14 have taken courses in both French and Russian 2092 students have taken a course in at least one of Spanish, French, and Russian How many students have taken a course in all 3 languages? Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 8
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Number of Elements in 3 Finite Sets Inclusion-Exclusion David O. Johnson EECS 210 (Fall 2016) 9 Solution : Let S be the set of students who have taken a course in Spanish, F the set of students who have taken a course in French, and R the set of students who have taken a course in Russian. Then, we have:| S | = 1232, | F | = 879, | R | = 114, | S F | = 103, | S R | = 23, | F R | = 14, and | S F R | = 23. Using the equation: | S F R | = | S |+ | F |+ | R | − | S F | − | S R | − | F R | + | S F R | We obtain: 2092 = 1232 + 879 + 114 −103 −23 −14 + | S F R |.
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  • Fall '12
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  • Natural number, Prime number, David O. Johnson EECS, O. Johnson EECS, David O. Johnson

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