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lecnotes4.3

# lecnotes4.3 - Lecture Notes Concepts of Mathematics(21-127...

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Lecture Notes, Concepts of Mathematics (21-127) Lecture 1, Recitation A–D, Spring 2008 4 Induction and the Binomial Theorem 4.3 The Binomial Theorem and Principles of Counting Definition 4.1 We say that a set A is finite iff it has n elements for some n N 0 . In this case we write # A = | A | = n . The number n is the cardinality of A . Example 4.1 We have # { a, b } = 2 # = 0 # { n Z | n 2 < 41 } = # {- 6 , - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 , 5 , 6 } = 13 Theorem 4.1 (The Sum Rule) If A and B are disjoint finite sets with # A = m and # B = n , then #( A B ) = m + n . For now we will accept this theorem as true, and we will be able to prove this later when we study functions. Note that if A and B are not disjoint, then # A + # B overcounts #( A B ) by counting each element of A B twice: Theorem 4.2 For finite sets A and B , #( A B ) = # A + # B - #( A B ) . (See Venn diagram) Note that if A and B are disjoint, then #( A B ) = # A + # B . You can use this to compute # A by examining the union of two disjoint subsets of A : # A = #( A B ) + #( A \ B ) for any set B . Example 4.2 During one week a total of 46 Concepts students went to Dr. Mackey to com- plain about their lecture instructor or their recitation instructor. Of these, 32 complained about the lecture instructor, and 20 complained about recitation. How many complained about both? #( L R ) = # L + # R - #( L R ) = 32 + 20 - 46 = 6 . 1

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Example 4.3 In a sample of 100 plastic cups, it was found that 8 cups had defect A , 10 cups had defect B , and 3 had both defects. How many had neither defect? We essentially want to find #( A B ) , which can be thought of as the total number of cups minus #( A B ) . Then we have 100 - (8 + 10 - 3) = 100 - 15 = 85 . Theorem 4.2 can be extended to three or more sets by the principle of inclusion and exclusion . The idea is that in counting the number of elements in the union of several sets by counting the number of elements in each set, we have included too many elements more than once; so some need to be excluded . When more than two sets are involved, the attempt at exclusion will subtract too many elements, so some need to be included. For three sets, the principle of inclusion and exclusion states that | A B C | = | A | + | B | + | C | - | A B | - | A C | - | B C | + | A B C | . Example 4.4 Among 40 first-time campers at Camp Forlorn one week, 14 fell into the lake during the week, 13 suffered from poison ivy, and 16 got lost trying to find the dining hall. Three had poison ivy and fell into the lake, 5 fell into the lake and got lost, 8 had poison ivy and got lost, and 2 experienced all three misfortunes. How many first-time campers got through the week unscathed? Let F be the set of campers who fell into the lake, let P be the set who got poison ivy, and let L be the set that got lost. Then we have | F P L | = | F | + | P | + | L | - | F P | - | F L | - | P L | + | F P L | = 14 + 13 + 16 - 3 - 5 - 8 + 2 = 29 . Thus there were 40 - 29 = 11 campers who got through the week unscathed.
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