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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) 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. Ch5-2 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 n1 n2 cannot be done at the same time. then there are n1+n2 ways to do either task. n1 + n2 ways 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? Ch5-3 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 n1 into two tasks. If there are n1 ways to do the n1 × n2 ways first task and n2 ways to do the second task after n2 the first task has been done, then there are n1 n2 ways to do the procedure. Ch5-4 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 1 x 100 x Ν Example 4 How many different bit strings are there of length seven? Sol: 1 2 3 4 5 6 7 □ □ □ □ □ □ □ → 27 種 ↑ ↑ ↑ ↑ ↑ ↑ ↑ 0,1 0,1 0,1 ...... 0,1 Ch5-5 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: a1 b1 f(a1)=? 可以是b1～ bn, 共n種 a2 f b2 f(a2)=? 可以是b1～ bn, 共n種 ． ． ． ． ： ． ． am bn f(am)=? 可以是b1～ bn, 共n種 ∴nm Ch5-6 Example 7 How many one-to-one functions are there from a set with m elements 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 # Ch5-7 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 Ch5-8 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 Vi be the number of variable names of length i. V1 =26 V2 =26．36 – 5 ∴26 + 26．36 – 5 different names. Ch5-9 ※ The Inclusion-Exclusion Principle (排容原理) A B = A B - A B A B Example 17 How many bit strings of length eight either start with a 1 bit or end with the two bits 00 ? Sol: １２３４５６７８ □□□□□□□□ ↑ ↑ ...... ① 1 0,1 0,1 → 共27種 ② . . . . . . . . . . . . 0 0 → 共26種 ③ 1 . . . . . . . . . . . 0 0 → 共25種 27 +26 -25 種 Ch5-10 ※ Tree Diagrams Example 18 How many bit strings of length four do not have two consecutive 1s ? Sol: 0 (0000) 0 1 (0001) 0 0 0 (0010) ∴ 8 bit strings 1 0 (0100) 1 0 1 (0101) 0 0 0 (1000) 1 1 (1001) 1 0 (1010) bit 1 bit 3 Exercise: 11, 17, 23, 27, 38, 39, 47, 53 Ch5-11 Ex 38. How many subsets of a set with 100 elements have more than one element ? Sol: 100 100 100 100 (1) 100 99 98 ... = = 2100 - 101 2 ( 2) a1 , a2 ,..., a100 Thm. 4 of §4.3 subset: □,□,..., □ 2100 - 101 放 放 放 空集合及 不放 不放 不放 只有1個元素的集合 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 n a1=an, a2=an-1, a3=an-2, … 2 2 種string. Ch5-12 §5.2 The Pigeonhole Principle (鴿籠原理) Theorem 1 (The Pigeonhole Principle) If k+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 the k 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. Ch5-13 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 N objects are placed into k boxes, then N there is at least one box containing at least k objects. e.g. 21 objects, 10 boxes there must be one box containing at least 21 = 3 objects. 10 Ch5-14 100 Example 5 Among 100 people there are at least 12 = 9 who were born in the same month. ( 100 objects, 12 boxes ) Ch5-15 Def. Suppose that a1 , a2 ,..., a N is a sequence of numbers. A subsequence of this sequence is a sequence of the form ai1 , ai2 ,..., aim where 1 i1 i2 ... im N (i.e., 保持原順序 ) 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 ai ai 1 A sequence is called decreasing (遞減) if ai ai 1 A sequence is called strictly increasing (嚴格遞增) if ai ai 1 A sequence is called strictly decreasing (嚴格遞減) if ai ai 1 Exercise: 5, 13, 15, 31 Ch5-16 §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 an r-permutation. Example 2. Let S = {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 with n distinct elements is P(n, r ) = n (n - 1) (n - 2)...( n - r 1) = n! 位置： 1 2 3 … r (n - r )! □ □ □ … □ 放法： n n -1 n - 2 … n - r 1 Ch5-17 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: P(100,3) = 100 99 98 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: 7!= 5040 Ch5-18 Def. An r-combination of elements of a set is an unordered selection of r elements from the set. Example 9 Let 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 0 r n, equals C = C (n, r ) n r = (n ) r = p ( n,r ) r! = n! r!( n-r )! 稱為 binomial coefficient pf : P(n, r ) = C (n, r ) r! Ch5-19 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. n! n! C (n, r ) = = = C (n, n - r ) r!(n - r )! (n - r )! (n - (n - r ))! 組合意義：選 r 個拿走,相當於是選 n - r 個留下. Ch5-20 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: C (9,3) C (11,4) Exercise: 3, 11, 13, 21, 33, 34. Ch5-21 §5.4 Binomial Coefficients (二項式係數) Example 1. ( x y)3 = ( x y)( x y)( x y) = ? x3 ? x 2 y ? xy2 y 3 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 ( 3 )種不同來源的 xy2 2 xy2 的係數 = ( 3 ) 2 ∴ ( x y ) = ( 0 ) x (1 ) x y ( 2 ) xy (3 ) y 3 3 3 3 2 3 2 3 3 Theorem 1. (The Binomial Theorem, 二項式定理) Let x,y be variables, and let n be a positive integer, then n ( x y) n = (0 ) x n (1 ) x n-1 y ... ( n-1 ) xyn-1 ( n ) y n = ( nj ) x n- j y j n n n n j =0 Ch5-22 Example 4. What is the coefficient of x12y13 in the expansion of (2 x - 3 y ) ? 25 Sol: (2 x - 3 y ) 25 = (2 x (-3 y )) 25 ∴ (13 ) 212 (-3)13 25 Cor 1. Let n be a positive integer. Then n k ( n ) = ( 0 ) (1 ) ... ( n ) = 2 n k =0 n n n pf : By Thm 1, let x = y = 1 (1 1) n = ( 0 ) (1 ) ( n ) ... ( n ) n n 2 n n Cor 2. Let n be a positive integer. Then ( -1) k ( n ) = 0 k k =0 pf : by Thm 1. (1-1)n = 0 Ch5-23 Theorem 2. (Pascal’s identity) Let n and k be positive integers with n k Then n 1 n n k = k - 1 k PASCAL’s triangle (0 ) 0 1 (1 ) 0 (1 ) 1 1 1 2 2 2 1 2 1 ( ) ( ) ( ) 0 1 2 1 3 3 1 (3 ) 3 (1 ) (3 ) (3 ) 1 4 6 4 1 0 2 3 … … 4 (3 ) Ch5-24 n 1 n n k = k - 1 k pf : ①(algebraic proof, 代數證明) n 1 = (n 1)! k k!(n 1 - k )! n n = n! n! k - 1 k (k - 1)! (n - k 1)! k!(n - k )! k n! (n - k 1) n! (n 1) n! = = k!(n - k 1)! k!(n - k 1)! k!(n - k 1)! ②(combinatorial proof, 組合意義證明): n 1 n 1 n 1 ‧ 取法 ‧ + ‧ n 1 = n 1 n 1 = k k - 1 0 k 1 k k 0 k-1 1 Ch5-25 Theorem 3. (Vandermode’s Identity) m, n, r , 0 r m, n r C (m n, r ) = C (m, r - k ) C (n, k ) k =0 pf : m n mn mn mn C(m+n, r) = = ↓ ↓ + ↓ ↓ +...+ ↓ ↓ r, 0 r-1, 1 0, r r = m n m n ... m n r 0 r - 1 1 0 r r = C (m, r - k ) C (n, k ) k =0 Ch5-26 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) Each path can be represented by a bit string consisting of m 0s and n 1s. 1 3 6 10 15 21 (→) (↑) 1 2 3 4 5 6 Red path對應的字串: 0 1 1 0 0 0 1 0 (0,0) 1 1 1 1 1 5 3 = 8! = 56 5 5!3! m n There are n paths of the desired type. Exercise: 7, 21, 28 Ch5-27

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