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Chapter 11

Chapter 11. Counting Methods. © 2008 Pearson Addison-Wesley. All rights reserved. Chapter 11: Counting Methods. 11.1 Counting by Systematic Listing 11.2 Using the Fundamental Counting Principle 11.3 Using Permutations and Combinations 11.4 Using Pascal’s Triangle

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Chapter 11

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  1. Chapter 11 Counting Methods © 2008 Pearson Addison-Wesley. All rights reserved

  2. Chapter 11: Counting Methods 11.1 Counting by Systematic Listing 11.2 Using the Fundamental Counting Principle 11.3 Using Permutations and Combinations 11.4 Using Pascal’s Triangle 11.5 Counting Problems Involving “Not” and “Or” © 2008 Pearson Addison-Wesley. All rights reserved

  3. Chapter 1 Section 11-5 Counting Problems Involving “Not” and “Or” © 2008 Pearson Addison-Wesley. All rights reserved

  4. Counting Problems Involving “Not” and “Or” • Problems Involving “Not” • Problems Involving “Or” © 2008 Pearson Addison-Wesley. All rights reserved

  5. Counting Problems Involving “Not” and “Or” The counting techniques in this section, which can be thought of as indirect techniques, are based on some correspondences between set theory, logic, and arithmetic as shown on the next slide. © 2008 Pearson Addison-Wesley. All rights reserved

  6. Set Theory/Logic/Arithmetic Correspondences © 2008 Pearson Addison-Wesley. All rights reserved

  7. Problems Involving “Not” Suppose U is the set of all possible results of some type. Let A be the set of all those results that satisfy a given condition. The figure below suggests that U A © 2008 Pearson Addison-Wesley. All rights reserved

  8. Complements Principle of Counting The number of ways a certain condition can be satisfied is the total number of possible results minus the number of ways the condition would not be satisfied. Symbolically, if A is any set within the universal set U, then © 2008 Pearson Addison-Wesley. All rights reserved

  9. Example: Counting the Proper Subsets of a Set For the set S = {c, a, l, u, t, o, r}, find the number of proper subsets. Solution A proper subset has less than seven elements. Subsets of many sizes would satisfy this condition. It is easier to consider the one subset that is not proper, namely S itself. S has a total of 27 = 128 subsets. From the complements principle, the number of proper subsets is 128 – 1 = 127. © 2008 Pearson Addison-Wesley. All rights reserved

  10. Example: Counting Coin Tossing If five fair coins are tossed, in how many ways can at least one tail be obtained? Solution By the fundamental counting principle, there are 25 = 32 different results possible. Exactly one of these fails to satisfy “at least one tail.” So from the complement principle we have the answer: 32 – 1 = 31. © 2008 Pearson Addison-Wesley. All rights reserved

  11. Problems Involving “Or” Another technique to count indirectly is to count the elements of a set by breaking that set into simpler component parts. If the cardinal number formula (Section 2.4) says to find the number of elements in S by adding the number in A to the number in B. We must then subtract the number in the intersection if A and B are not disjoint. If A and B are disjoint, the subtraction is not necessary. © 2008 Pearson Addison-Wesley. All rights reserved

  12. Additive Principle of Counting The number of ways that one or the other of two conditions could be satisfied is the number of ways one of them could be satisfied plus the number of ways the other could be satisfied minus the number of ways they could both be satisfied together. If A and B are any two sets, then If A and B are disjoint, then © 2008 Pearson Addison-Wesley. All rights reserved

  13. Example: Counting Card Hands How many five-card hands consist of all hearts or all black cards? Solution The sets all hearts and all black cards are disjoint. n(all hearts or all black cards) = n(all hearts) + n(all black cards) = 13C5 + 26C5 = 1,287 + 65,780 = 67,067 © 2008 Pearson Addison-Wesley. All rights reserved

  14. Example: Counting Selections From a Diplomatic Delegation A diplomatic delegation of 20 congressional members are categorized as to political party and gender. If one of the members is chosen randomly to be spokesperson for the group, in how many ways could that person be a Republican or a man? © 2008 Pearson Addison-Wesley. All rights reserved

  15. Example: Counting Selections From a Diplomatic Delegation Solution © 2008 Pearson Addison-Wesley. All rights reserved

  16. Example: Counting Three-Digit Numbers with Conditions How many three-digit counting numbers are multiples of 2 or multiples of 5? Solution There are 9(10)(5) = 450 three-digit multiples of 2. A multiple of 5 ends in a 0 or 5, so there are 9(10)(2) = 180 of those. A multiple of 2 and 5 must end in a 0. There are 9(10)(1) = 90 of those. So we have 450 + 180 – 90 = 540 three-digit counting numbers that are multiples of 2 or 5. © 2008 Pearson Addison-Wesley. All rights reserved

  17. Example: Counting Card-Drawing Results A single card is drawn from a standard 52-card deck. a) In how many ways could it be a club or a queen? b) In how many ways could it be a red card or a face card? Solution a) club + queen – queen of clubs = 13 + 4 – 1 = 16? b) red card + face card – red face cards = 26 + 12 – 6 = 32 © 2008 Pearson Addison-Wesley. All rights reserved

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