12月3日旅游与酒店管理学院酒店Q0441.pptx

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1、Discrete Mathematics Chapter 5 Counting大葉大學大葉大學 資訊工程系資訊工程系 黃鈴玲黃鈴玲Ch5-2 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 ma

2、ny possible passwords are there? This section introducesa variety of other counting problemsthe basic techniques of counting.5.1 The Basics of countingCh5-3Basic 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 a

3、t the same time. then there are n1+n2 ways 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?n1

4、n2n1 + n2 waysCh5-4Example 12 A student can choose a computer project from one of three lists. The three lists contain23, 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 d

5、own into two tasks. If there are n1 ways to do thefirst task and n2 ways to do the second task after the first task has been done, then there aren1 n2 ways to do the procedure.n1n2n1 n2 waysCh5-5Example 2 The chair of an auditorium (大禮堂) is to be labeled with a letter and a positive integer not exce

6、eding 100. What is the largest number of chairs that can be labeled differently?Sol: 26 100 = 2600 ways to label chairs. letter xx1001Example 4 How many different bit strings are there of length seven?Sol:12 3 4 5 6 7 0,1 0,1 0,1 . . . . . . 0,1 27 種Ch5-6Example 5 How many different license plates (

7、車牌) are available if each plate contains a sequence of 3 letters followed by 3 digits ?Sol: 263103 letter digitExample 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 nma1a2amb1b2bnfCh5-7Example

8、 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-

9、1 function #Ch5-8Example 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,

10、7,8 P6 = 366 - 266 P7 = 367 - 267 P8 = 368 - 268 P6 + P7 + P8 = 366 + 367 + 368 - 266 - 267 - 268種Ch5-9Example 14 In a version of Basic, the name of a variableis a string of one or two alphanumeric characters, whereuppercase and lowercase letters are not distinguished.Moreover, a variable name must

11、begin with a letter andmust be different from the five strings of two charactersthat 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 =2636 5 26 + 2636 5 different names.Ch5-10 Th

12、e Inclusion-Exclusion Principle (排容原理排容原理) A BExample 17 How many bit strings of length eight either start with a 1 bit or end with the two bits 00 ?Sol: . . . . . . 1 0,1 0,1 共27種 . . . . . . . . . . . . 0 0 共26種 1 . . . . . . . . . . . 0 0 共25種 27 +26 -25 種BABABA-=Ch5-1101bit 1 Tree Diagrams Examp

13、le 18 How many bit strings of length four do not have two consecutive 1s ? Sol: Exercise: 11, 17, 23, 27, 38, 39, 47, 5300000111001(0000)(0001)(0010)(0100)(0101)(1000)(1001)(1010) 8 bit strings00110bit 3Ch5-12Ex 38. How many subsets of a set with 100 elements have more than one element ? Sol: Ex 39.

14、 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, 1012 2100.9810099100100100 ) 1 (100-=Thm. 4 of 4.310021,., )2(aaa1012 ,., , :subset 10

15、0- 放不放 放不放 放不放空集合及只有1個元素的集合string.種22nCh5-13 5.2 The Pigeonhole Principle (鴿籠原理鴿籠原理)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

16、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.Ch5-14Example 2 In any group of 27 English words, there must be at least two that beg

17、in 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 ? (0100 points)Sol: 102. (101+1)Theorem 2. (The generalized pigeon hole principle) If N objects are placed into k boxes, then there is at least one

18、box containing at least objects.e.g. 21 objects, 10 boxes there must be one box containing at least objects.kN31021=Ch5-15Example 5 Among 100 people there are at least who were born in the same month. ( 100 objects, 12 boxes )912100=Ch5-16Example 10 During a month with 30 days a baseball team plays

19、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.45sum 存在一段時間的game數和=14(跳過)Ch5-17Sol:Let aj be the number of games played on or before thejth day of the month. (第1天第j天的比賽數和)Then is a

20、n increasing sequence of distinct integers withMoreover, is also an increasingsequence of distinct integers withThere are 60 positive integersbetween 1 and 59. Hence, such that)301 ( j3021,.,aaajaj 451)45.1 .,.(30321aaaaei14,.,14,143021aaa591415ja)5914.14141415 .,.(30321aaaaei14,.,14,.,301301aaaaji

21、and )場14共 天天第1第 .,.( 14ijeiaaji=#(跳過)Ch5-18Def. 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 calle

22、d decreasing (遞減) if A sequence is called strictly increasing (嚴格遞增) if A sequence is called strictly decreasing (嚴格遞減) if Naaa,.,21miiiaaa,.,21Niiim.121) .,.(保持原順序ei1iiaa1iiaa1iiaa1iiaaCh5-19Theorem 3. Every sequence of n2+1 distinct real numbers contains a subsequence of length n+1 that is either

23、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 C

24、onstruct a sequence of 16 positive integers that has no increasing or decreasing subsequence of 5 terms.Sol:Exercise: 5, 13, 15, 31(跳過)13,14,15,16 9 ,10,11,12 5 ,6,7,8 1 ,2, 3 ,4 Ch5-20 5.3 Permutations(排列排列) and Combinations(組合組合)Def. A permutation of a set of distinct objects is an ordered arrange

25、ment 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 位置： 1 2 3 r

26、 放法：)!(!) 1).(2() 1(),(rnnrnnnnrnP-=-=n1-n2-n1-rnCh5-21Example 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 diff

27、erent 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:9899100) 3 ,100(=P5040! 7 =Ch5-22Def. An r-combination of elements of a set is an unordered selection

28、 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 , equals pf :)!( !),()(),(rnrnrrnpnrrnCCnr-=nr 0!),(),(rrnCrnP=稱為 binomial

29、 coefficientCh5-23Example 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,dCorollary 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.),()!()!(!)!( !),(rnnCrnnrnnrnrnrn

30、C-=-=-=Ch5-24Example 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)=252Example 15. Suppose there are 9 faculty members in the math department and 11 in the computer science department. How many ways are there to

31、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.)4 ,11()3 , 9(CCCh5-25 5.4 Binomial Coefficients (二項式係數二項式係數)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

32、跟 y 不可能相乘) 共有 種不同來源的 xy2 32233?)()()(yxyyxxyxyxyxyx=)(32)(323332322313303)()()()()(yxyyxxyx=-=njjjnnjnnnnnnnnnnnyxyxyyxxyx011110)()()(.)()()( 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, thenCh5-26Example 4. What is the coefficient of x12y1

33、3 in the expansion of ?Sol: Cor 1. Let n be a positive integer. Thenpf : By Thm 1, let x = y = 1 Cor 2. Let n be a positive integer. Thenpf : by Thm 1. (1-1)n = 025)32(yx-2525)3(2()32(yxyx-=-13122513)3(2)(-)(.)()()() 11 (210nnnnnn=-nknkk00)() 1(=nknnnnnnk0102)(.)()()(Ch5-27Theorem 2. (Pascals identi

34、ty) Let n and k be positive integers with n k Then-= knknkn11PASCALs triangle)(00)(10)(11)(20)(21)(22)(30)(31)(32)(33)(43111121133141461Ch5-28pf : (algebraic proof, 代數證明)n 1k取法 =1kn 01k-1+n 1-= knknkn11(combinatorial proof, 組合意義證明):)!1( !)!1(1knknkn-= )!( !)!1()!1(!1knknknknknkn-=-)!1( !) 1()!1( !)

35、1()!1( !-=-=knknnknknknknknk-= 110111knknknCh5-29Theorem 3. (Vandermodes Identity)pf :=-=rkknCkrmCrnmCnmrrnm0),(),(),(,0 ,=-=rkknCkrmC0),(),(mnrC(m+n, r) = m n m n m n + +.+ r, 0 r-1, 0 0, r=-=rnmnrmnrm0.110Ch5-30Ex 33. Here we will count the number of paths between the origin (0,0) and point (m,n)

36、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 1Red path對應的字串: 0 1 1 0 0 0 1 0 Each path can be represented by a bit string consisting of m 0s and n 1s. () ()56 ! 3 !5!8535=There are paths of the desired type.nnmExercise: 7, 21, 28