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Discrete Mathematics

Discrete Mathematics. Chapter 5 Counting. 大葉大學 資訊工程系 黃鈴玲. §5.1 The Basics of counting.  A counting problem: (Example 15)

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Discrete Mathematics

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  1. Discrete Mathematics Chapter 5 Counting 大葉大學 資訊工程系 黃鈴玲

  2. §5.1 The Basics of counting  A counting problem: (Example 15) Each user on a computer system has a password, which is six to eight characters long, where each characters is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there?  This section introduces • a variety of other counting problems • the basic techniques of counting.

  3. n1 n2 n1 + n2 ways Basic counting principles  The sum rule: If a first task can be done in n1 ways and a second task in n2 ways, and if these tasks cannot be done at the same time. then there are n1+n2ways to do either task. Example 11 Suppose that either a member of faculty or a student is chosen as a representative to a university committee. How many different choices are there for this representative if there are 37 members of the faculty and 83 students?

  4. n1 n1 × n2 ways n2 Example 12 A student can choose a computer project from one of three lists. The three lists contain 23, 15 and 19 possible projects respectively. How many possible projects are there to choose from? Sol:23+15+19=57 projects.  The product rule: Suppose that a procedure can be broken down into two tasks. If there are n1 ways to do the first task and n2 ways to do the second task after the first task has been done, then there are n1 n2 ways to do the procedure.

  5. Example 2 The chair of an auditorium (大禮堂) is to be labeled with a letter and a positive integer not exceeding 100. What is the largest number of chairs that can be labeled differently? Sol:26 × 100 = 2600 ways to label chairs. letter Example 4 How many different bit strings are there of length seven? Sol:123456 7 □ □ □ □ □ □ □ ↑ ↑ ↑ ↑ ↑ ↑ ↑ 0,1 0,1 0,1 . . . . . . 0,1 → 27種

  6. a1 a2 ． ． ． am b1 b2 ． ． ． bn f Example 5 How many different license plates (車牌) are available if each plate contains a sequence of 3 letters followed by 3 digits ? Sol: □ □ □ □ □ □ →263．103 letter digit Example 6 How many functions are there from a set with m elements to one with n elements? Sol:f(a1)=?可以是b1～ bn, 共 n種 f(a2)=?可以是b1～ bn, 共 n種 ： f(am)=?可以是b1～ bn, 共 n種 ∴nm

  7. Example 7 How many one-to-one functions are there from a set with melements to one with n element? (m  n) Sol:f(a1) = ?可以是b1～ bn, 共 n 種 f(a2) = ?可以是b1～ bn, 但不能=f(a1), 共 n-1 種 f(a3) = ?可以是b1～ bn, 但不能=f(a1), 也不能=f(a2), 共 n-2 種 ： ： f(am) = ?不可=f(a1), f(a2), ... , f(am-1), 故共n-(m-1)種 ∴共 n．(n-1)．(n-2)．...．(n-m+1)種 1-1 function #

  8. Example 15 Each user on a computer system has a password which is 6 to 8 characters long, where each character is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there? Sol:Pi :# of possible passwords of length i , i=6,7,8 P6 = 366 - 266 P7 = 367 - 267 P8 = 368 - 268 ∴ P6+ P7+ P8= 366 + 367 + 368 -266- 267 - 268種

  9. Example 14 In a version of Basic, the name of a variable is a string of one or two alphanumeric characters, where uppercase and lowercase letters are not distinguished. Moreover, a variable name must begin with a letter and must be different from the five strings of two characters that are reserved for programming use. How many different variable names are there in this version of Basic? Sol: Let Vibe the number of variable names of length i. V1 =26 V2 =26．36 – 5 ∴26 + 26．36 – 5 different names.

  10. ※ The Inclusion-Exclusion Principle (排容原理) A B Example 17How many bit strings of length eight either start with a 1 bit or end with the two bits 00 ? Sol: １ ２ ３ ４ ５ ６ ７ ８ □ □ □ □ □ □ □ □ ↑ ↑ . . . . . . ① 10,1 0,1 → 共27種 ② 　. . . . . . . . . . . . 0 0 → 共26種 ③ 1 . . . . . . . . . . . 0 0 → 共25種  27 +26 -25 種

  11. (0000) 0 0 1 (0001) 0 (0010) 0 ∴ 8 bit strings 0 1 (0100) 0 1 0 (0101) 1 0 0 (1000) 0 1 1 (1001) 1 (1010) 0 bit 1 bit 3 ※ Tree Diagrams Example 18 How many bit strings of length four do not have two consecutive 1s ? Sol: Exercise: 11, 17, 23, 27, 38, 39, 47, 53

  12. Ex 38. How many subsets of a set with 100 elements have more than one element ? Sol: Ex 39. A palindrome (迴文) is a string whose reversal is identical to the string. How many bit strings of length n are palindromes ? ( abcdcba 是迴文, abcd 不是 ) Sol: If a1a2... an is a palindrome, then a1=an, a2=an-1,a3=an-2, … Thm. 4 of §4.3 空集合及只有1個元素的集合 放 不放 放 不放 放 不放

  13. §5.2 The Pigeonhole Principle (鴿籠原理) Theorem 1 (The Pigeonhole Principle) Ifk+1 or more objects are placed into k boxes, then there is at least one box containing two or more of the objects. Proof Suppose that none of thek boxes contains more than one object. Then the total number of objects would be at most k. This is a contradiction. Example 1. Among any 367 people, there must be at least two with the same birthday, because there are only 366 possible birthdays.

  14. Example 2 In any group of 27 English words, there must be at least two that begin with the same letter. Example 3 How many students must be in a class to guarantee that at least two students receive the same score on the final exam ? (0~100 points) Sol:102. (101+1) Theorem 2. (The generalized pigeon hole principle) If Nobjects are placed into k boxes, then there is at least one box containing at least objects. e.g. 21 objects, 10 boxes  there must be one box containing at least objects.

  15. Example 5 Among 100 people there are at least who were born in the same month. ( 100 objects, 12 boxes )

  16. (跳過) Example 10 During a month with 30 days a baseball team plays at least 1 game a day, but no more than 45 games. Show that there must be a period of some number of consecutive days during which the team must play exactly 14 games. 存在一段時間的game數和=14

  17. (跳過) Sol: Let ajbe the number of games played on or before the jth day of the month. (第1天～第j天的比賽數和) Then is an increasing sequence of distinct integers with Moreover, is also an increasing sequence of distinct integers with There are 60 positive integers between 1 and 59. Hence, such that

  18. Def.Suppose that is a sequence of numbers. A subsequence of this sequence is a sequence of the form where e.g. sequence: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7 subsequence: 8, 9, 12 () 9, 11, 4, 6 () Def. A sequence is called increasing(遞增) if A sequence is calleddecreasing(遞減) if A sequence is called strictly increasing(嚴格遞增)if A sequence is calledstrictly decreasing(嚴格遞減)if

  19. (跳過) Theorem 3. Every sequence of n2+1distinct real numbers contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing. Example 12. The sequence 8, 11, 9, 1, 4, 6, 12, 10, 5, 7 contains 10=32+1 terms (i.e., n=3). There is a strictly increasing subsequence of length four, namely, 1, 4, 5, 7. There is also a decreasing subsequence of length 4, namely, 11, 9, 6, 5. Exercise 21 Construct a sequence of 16 positive integers that has no increasing or decreasing subsequence of 5 terms. Sol: Exercise: 5, 13, 15, 31

  20. §5.3 Permutations(排列) and Combinations(組合) Def. A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement of r elements of a set is called anr-permutation. Example 2.LetS ={1, 2, 3}.The arrangement 3,1,2 is a permutation of S. The arrangement 3,2 is a 2-permutation of S. Theorem 1. The number of r-permutation of a set withn distinct elements is 位置： 1 2 3 … r 　　　□ □ □ … □ 放法： …

  21. Example 4. How many different ways are there to select a first-prize winner (第一名), a second-prize winner, and a third-prize winner from 100 different people who have entered a contest ? Sol: Example 6. Suppose that a saleswoman has to visit 8 different cities. She must begin her trip in a specified city, but she can visit the other cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities ? Sol:

  22. 稱為 binomial coefficient Def. An r-combination of elements of a set is an unordered selection of r elements from the set. Example 9Let S be the set {1, 2, 3, 4}. Then {1, 3, 4} is a 3-combination from S. Theorem 2 The number of r-combinations of a set with n elements, where n is a positive integer and r is an integer with , equals pf :

  23. Example 10. We see that C(4,2)=6, since the 2-combinations of {a,b,c,d} are the six subsets {a,b}, {a,c}, {a,d}, {b,c}, {b,d} and {c,d} Corollary 2. Let n and r be nonnegative integers with r  n. Then C(n,r) = C(n,n-r) pf : From Thm 2. 組合意義：選 r 個拿走,相當於是選 n - r 個留下.

  24. Example 12. How many ways are there to select 5 players from a 10-member tennis team to make a trip to a match at another school ? Sol: C(10,5)=252 Example 15. Suppose there are 9 faculty members in the math department and 11 in the computer science department. How many ways are there to select a committee if the committee is to consist of 3 faculty members from the math department and 4 from the computer science department? Sol: Exercise:3, 11, 13, 21, 33, 34.

  25. §5.4 Binomial Coefficients (二項式係數) Example 1. 要產生 xy2項時， 需從三個括號中選兩個括號提供 y，剩下一個則提供 x (注意：同一個括號中的 x跟 y不可能相乘) ∴共有 種不同來源的 xy2 ∴  xy2的係數 = Theorem 1.(The Binomial Theorem, 二項式定理) Let x,y be variables, and let n be a positive integer, then

  26. Example 4. What is the coefficient of x12y13 in the expansion of ? Sol: ∴ Cor 1. Let n be a positive integer. Then pf : By Thm 1, let x = y = 1 Cor 2. Let n be a positive integer. Then pf : by Thm 1. (1-1)n = 0

  27. … … Theorem 2.(Pascal’s identity) Let nand k be positive integers with n k Then PASCAL’s triangle

  28. n 1 n n 1 1 + ‧ ‧ ‧ k k 0 k-1 1 pf : ①(algebraic proof, 代數證明) ②(combinatorial proof, 組合意義證明): 取法= 

  29. m n r Theorem 3.(Vandermode’s Identity) pf : C(m+n, r) = m n m n m n ↓ ↓ + ↓ ↓ +...+ ↓ ↓ = r,0 r-1, 1 0, r

  30. Ex 33. Here we will count the number of paths between the origin (0,0) and point (m,n) such that each path is made up of a series of steps, where each step is a move one unit to the right or a more one unit upward. 1 4 10 20 35 56 (5,3) 1 3 6 10 15 21 1 2 3 4 5 6 (0,0) 1 1 1 1 1 Each path can be represented by a bit string consisting of m0s and n1s. (→) (↑) Red path對應的字串: 0 1 1 0 0 0 1 0 There are paths of the desired type. Exercise: 7, 21, 28

  31. §5.5 Generalized Permutations and Combinations Permutations with Repetition Example 1 How many strings of length r can be formed from the English alphabet? Sol. 26r Thm 1. The number of r-permutations of a set of n objects with repetition allowed is nr.

  32. Combinations with Repetition Example 3 How many ways are there to select five bills from a cash box containing $1 bills, $2 bills, $5 bills, $10 bills, $20 bills, $50 bills, and $100 bills? (假設順序不計、同面額不區分、每種面額至少五張) Sol. 的個數表示張數 隔開的bar  可用字串 |  |  | |  |  | 代表  可用字串   |  | |   | | | 代表  可用字串    | | |  | | |  代表 … 7種面額需6個bar隔開，故6個bar 跟5張bill組合  C(11, 5)

  33. a3不出現 a2出現1次 取法 a1出現2次 a4出現3次 ∴ 種方法 # Thm 2. There are C(r+n-1, r) r-combinations from a set with nelements when repetition of elements is allowed. pf : (視為有 r個＊，要放入 n 個區域，故需插入 n-1個 bar 將這些＊隔成n 區) 例： 設n = 4, 集合為 , r = 6 ＊＊｜＊｜｜＊＊＊

  34. Example 4. Suppose that a cookie shop has 4 different kinds of cookies. How many different ways can 6 cookies be chosen? Sol:6個cookie插入3個bar 種 Example 5. How many solutions does the equation x1 + x2 + x3 = 11 have, where x1, x2, x3 are nonnegative integers?’ Sol:11個1間要插入2個bar 種

  35. ※若上題改要求 x1  1, x2  2, x3  3, 則原式 x1 + x2 + x3 = 11 可改成 (x1 -1) + (x2 -2)+ (x3 -3) = 11 - 1 - 2 - 3 = 5 此時相當於是 y1 + y2 + y3 = 5 其中 y1 = x1 -1, y2 = x2 -2, y3 = x3 -3 且 y1, y2 , y3 N ∴5個1間要插入2個bar  種 (注意： case y1 = 1, y2 = 3, y3 = 1 相當於 x1 =2, x2 =5, x3 =4) ※若上題再改為 1 x1  3, x2  2, x3  3，則需排除 x1 > 3的情形(即x1  4的情形) 因 (x1 -4) + (x2 -2)+ (x3 -3) = 11 - 9 = 2 ∴共有 種 Exercise: 15, 20, 25

  36. ※ Permutations with indistinguishable objects Example 7. How many different strings can be made by reordering the letters of the word SUCCESS ? Sol: 有3個S, 2個C, 1個U 及 1個E, 可放S的位置有 種 剩下的4個位置中可放C的有 種 剩下的2個位置中可放U的有 種 剩下的1個位置中可放E的有 種 ∴共 種

  37. Thm 3.The number of different permutations of n objects, where type 1： n1 種 type 2： n2 種 is type k： nk 種 pf : Exercise: 17, 31, 36, 65

  38. Distributing objects into Boxes Distinguishable Objects and distinguishable boxes Example 8. How many ways are there to distribute hands of 5 cards to each of four players from the standard deck of 52 cards? Sol: player 1： 種 player 2： 從剩下的牌再發5張  注意：上題相當於將52張牌放進5個不同的box的放法, 即 box 1 的給 player 1 box 2 的給 player 2 box 3 的給 player 3 box 4 的給 player 4 　　　而 box 5 的是剩下的牌. Exercise: 45

  39. Thm 4. The number of ways to distribute n distinguishable objects into k distinguishable boxes so that ni objects are placed into box i, i=1, 2, …, k, equals (跟Thm 3相等) Indistinguishable Objects and distinguishable boxes Example 9. How many ways are there to place 10 indistinguishable balls into eight distinguishable bins? Sol: C(8+10-1, 10) There are C(n+r-1, n-1) ways to place r indistinguishableobjects into n distinguishable boxes.

  40. Distinguishable Objects and indistinguishable boxes Example 10. How many ways are there to put four different employees into three indistinguishable offices, when each office can contain any number of employees? Sol: employees: A, B, C, D 4人在同一office: {{A, B, C, D}} 1種 3人在同一office, 1人在另一office: {{A, B, C}, {D}}, {{A, B, D}, {C}},… 4種 2人在同一office, 2人在另一office: {{A, B}, {C, D}}, {{A, D}, {B, C}},… 3種 2人在同一office,另2人分別在剩下兩個office: {{A, B}, {C}, {D}}, {{A, D}, {B}, {C}},… 6種 共 14種 Note. There is no simple closed formula. 可參考 Stirling numbers of the second kind. (p. 378)

  41. Indistinguishable Objects and indistinguishable boxes Example 11. How many ways are there to pack six copies of the same book into four identical boxes, where a box can obtain as many as six books? Sol: 2, 2, 1, 1 3, 3 6 3, 2, 1 5, 1 共 9種 3, 1, 1, 1 4, 2 2, 2, 2 4, 1, 1 Note. This problem is the same as writing n as the sum of atmost k positive integers in nonincreasing order.That is, a1+a2+…+aj = n, where a1, a2, …, ajare positive integers with a1  a2  …  aj and j  k. No simple closed formula exists.

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