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- 3/26/03 Tucker, Applied Combinatorics Section 5.11 5.1 General Counting Methods for Arrangements and Selections Tucker, Applied Combinatorics, Sec. 5.1,

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Slide 1 3/26/03 Tucker, Applied Combinatorics Section 5.11 5.1 General Counting Methods for Arrangements and Selections Tucker, Applied Combinatorics, Sec. 5.1, Patti Bodkin and Tamsen Hunter Slide 2 3/26/03 Tucker, Applied Combinatorics Section 5.12 Principals The Addition Principal If there are different objects in the first set, objects in the second set,..., and the objects in the mth set, and if the different sets are disjoint, then then number of the ways to select an object from one of the m sets is. Slide 3 3/26/03 Tucker, Applied Combinatorics Section 5.13 Principals The Multiplication Principal Suppose a procedure can be broken into m successive (ordered) stages, with outcomes in the first stage, outcomes in the second stage,..., and outcomes in the mth stage. If the number of outcomes at each stage is independent of the choices in previous stages and if the composite outcomes are all distinct, then the total procedure has different composite outcomes. Slide 4 3/26/03 Tucker, Applied Combinatorics Section 5.14 Example 1 There are: five different Spanish books six different French books eight different Transylvanian books. How many books are there to pick an (unordered) pair of two books not both in the same language? If one Spanish and one French book are chosen, using the multiplication principal, the selection can be done; 5X6=30. If one Transylvanian book and one Spanish book, then 5X8=40. If one Transylvanian and one French book, then 6X8=48 ways. These three types of selections are disjoint, so by the addition principal there are 30+40+48=118 ways in all. Slide 5 3/26/03 Tucker, Applied Combinatorics Section 5.15 Tip for Solving Problems The preceding example typifies a basic way of thinking in combinatorial problem solving: Always try first to break a problem into moderate number of manageable subproblems. There may be cleverer ways, but if we can reduce the original problem to subproblems with which we are familiar, then we are less likely to make a mistake. Slide 6 3/26/03 Tucker, Applied Combinatorics Section 5.16 Example 2 How many ways are there to form a three-letter sequence using the letters a, b, c, d, e, f: 1: with repetition of letters allowed? 2: without repetition of any letter? 3: without repetition and containing the letter e? 4: with repetition and containing the letter e? Slide 7 3/26/03 Tucker, Applied Combinatorics Section 5.17 Example 2.1: How many 3-letter sequences with repetition allowed? With repetition, we have six choices for each letter in the sequence. So by the multiplication principle there are three-letter sequences without repetition. a a b Slide 8 3/26/03 Tucker, Applied Combinatorics Section 5.18 Example 2.2: How many 3-letter sequences without repetition? Without repetition, there are six choices for the first letter. For the second letter, there are five choices. Similarly for the third letter, there are four choices. Thus there are three- letter sequences without repetition. f d b Slide 9 3/26/03 Tucker, Applied Combinatorics Section 5.19 Example 2.3: How many 3-letter sequences without repetition and containing the letter e? Use a diagram to display the positions in a sequence. Since the sequence must contain an e, then there are three choices for which position in the sequence is e: eee Slide 10 3/26/03 Tucker, Applied Combinatorics Section 5.110 Example 2.3 (continued) How many 3-letter sequences without repetition and containing the letter e? In each diagram, there are 5 choices for which of the other 5 letters (excluding e) goes in the first remaining position and 4 choices for which of the remaining 4 letters goes in the other position. Thus there are three-letter sequences with e. edc Slide 11 3/26/03 Tucker, Applied Combinatorics Section 5.111 Example 2.4: How many 3-letter sequences with repetition and containing the letter e? As in part 3 when repetition is allowed, there are three choices for es position. For any of these choices for es position, there are choices for the other two positions, since e and the other letters can appear more than once. But the answer of is not correct. The Multiplication Principle has been violated because the outcomes are not distinct. Slide 12 3/26/03 Tucker, Applied Combinatorics Section 5.112 Example 2.4 (continued) How many 3-letter sequences with repetition and containing the letter e? Consider the sequence: It was generated two times in our procedure: once when e was put in the first position followed by c e as one of the 36 choices for the latter two positions, and a second time when e was put in the last position with e c in the first two positions. We must use an approach that ensures distinct outcomes. Let us break the problem into disjoint cases based on where the first e in the sequence occurs. e c e Slide 13 3/26/03 Tucker, Applied Combinatorics Section 5.113 Example 2.4 (continued) How many 3-letter sequences with repetition and containing the letter e? First suppose the first e is in the first position: Then there are six choices (including e) for the second and for the third positions6 * 6 ways. Next suppose the first e is in the second position: Then there are five choices for the first position (cannot be e) and six choices for the last position5 * 6 ways. Finally, let the first (and only) e be in the last position: There are five choices each for the two positions5 * 5 ways. The correct answer is thus: e a b e df e c d Slide 14 3/26/03 Tucker, Applied Combinatorics Section 5.114 Problem for Class to Try: How many nonempty different collections can be formed from five (identical) apples and eight (identical) oranges? Hint: Concentrate on what makes one collection different from another collection. Slide 15 3/26/03 Tucker, Applied Combinatorics Section 5.115 Problem for Class to Try: The number of apples and the number of oranges will be different in different collections. We can characterize any collections by a pair for integers (a,o), where a is the number of apples and o is the number of oranges. There are 6 possible values for a and 9 possible values for o. Together there are. Since the problem asked for non-empty collections and one of the solutions is (0,0), the desired answer is Slide 16 3/26/03 Tucker, Applied Combinatorics Section 5.116 Advice: When you are stuck and cannot get started with a problem, try writing down in a systematic fashion of the possible outcomes you want to enumerate. By doing this, you should start to see a pattern emerge. For help with problem # 45 on the homework, this link is good: http://www.cut-the-knot.com/SimpleGames/FrogsAndToads.shtml