Discrete Mathematics

Chapter 5 Counting

§5.1 The Basics of counting

We will present two basic counting principles, the product rule and the sum rule. The product rule: Suppose that a procedure can be broken down 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.

n1

n2

n1 × n2

ways

Example 1 A new company with just two employees, Sanchez and Patel, rents a floor of a building with 12 offices. How many ways are there to assign different offices to these two employees?

Sol: 12*11 =132 ways

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

x

x 1001

Example 4 How many different bit strings are there of length seven?Sol: 　 1 　 2 　 3 　 4 　 5 　 6 　 7 □ □ □ □ □ □ □ ↑ ↑ ↑ ↑ ↑ ↑ ↑ 0,1 0,1 0,1 . . . . . . 0,1

→ 27 種

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

a1

a2

．

．

．

am

b1

b2

．

．

．

bn

f

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 n1 種 f(a3) = ? 可以是 b1 ～ bn, 但不能 = f(a1), 也不能 =f(a

2), 共 n2 種 ： ： f(am) = ? 不可 f(a1), f(a2), ... , f(am1), 故共 n(m1) 種 ∴ 共 n ． (n1) ． (n2) ． ... ． (nm+1) 種 1-1 function

#

A password on a computer system consists of 6,7, or 8 characters. Each of these characters must be a digit or a letter of the alphabet. Each password must contain at least one digit. How many such passwords are there?

This section introduces- a variety of other counting problems- the basic techniques of counting.

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+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?

Sol: 37+83=120

n1 n2

n1 + n2 ways

Example 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.

Example 14 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

Example 13 In a version of Basic, the name of a variableis a string of one (only alphanumeric characters )or two alphanumeric characters, (An alphanumeric characters is either one of the 26 English letters or one of the 10 digits) 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.

※ The Inclusion-Exclusion Principle (排容原理 )

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: １ ２ ３ ４ ５ ６ ７ ８ □ □ □ □ □ □ □ □ ↑ ↑ . . . . . . 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

BABABA

0

1

bit 1

※ Tree Diagrams

Example 19 How many bit strings of length four do not have two consecutive 1s ? Sol:

Exercise: 11, 17, 21, 27, 36, 37, 45, 49

0

0

0

0

0

1

1

10

0

1

(0000)

(0001)

(0010)

(0100)(0101)(1000)

(1001)(1010)

∴ 8 bit strings

0

0

1

1

0

bit 3

Ex 36. How many subsets of a set with 100 elements have more than one element ? Sol:

Ex 37. A palindrome ( 迴文 ) is a string whose reversal is identical to the string. How many bit strings of length n are palindromes ? (e.g., abcdcba 是迴文 , abcd 不是 )

Sol: If a1a2 ... an is a palindrome, then a1=an, a2=an1, a3=an2, …

1012 2

100...

98

100

99

100

100

100 )1( 100

Thm. 4 of §5.3 10021 ,...,, )2( aaa

1012 □ ,..., □, □ :subset 100 放不放

放不放

放不放

空集合及只有 1 個元素的集合

string.種2 2

n

§5.2 The Pigeonhole Principle (鴿籠原理 )

Thm 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.

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)

Thm 2. (The generalized pigeon hole principle) If N objects 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.

k

N

310

21

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

912

100

Example 6 What is the minimum number of students required in a D.M class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D and E?Sol: [N/5] = 6; N= 5*5+1 =26

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.

day 1 2 3 4 5 ... 15 30

# of game 3 2 1 2 45sum

存在一段時間的 game 數和 =14

Some Elegant Applications of the Pigeonhole Principle

Sol:

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 an increasing sequence of distinct integers with

Moreover, is also an increasingsequence of distinct integers with

There are 60 positive integersbetween 1 and 59. Hence, such that

)301( j

3021 ,...,, aaaja j 451

)45...1 .,.( 30321 aaaaei

14,...,14,14 3021 aaa591415 ja

)5914...14141415 .,.( 30321 aaaaei

14,...,14,,..., 301301 aaaaji and

)場14共 天天～第1第 .,.( 14 ijeiaa ji #

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 called 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 ,...,, 21

miii aaa ,...,,21

Niii m ...1 21

) .,.( 保持原順序ei

1 ii aa

1 ii aa

1 ii aa1 ii aa

Thm 3. Every sequence of n2+1 distinct 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 (參考習題： 21, 23)13,14,15,16 9,10,11,12 5,6,7,8 1,2,3,4

§4.3 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 1. Let The arrangement 3,1,2 is a permutation of S. The arrangement 3,2 is a 2-permutation of S.

Thm 1. The number of r-permutation of a set with n distinct elements is 位置： 1 2 3 … r　　　□ □ □ … □　　　↑ ↑ ↑ … ↑放法：

}.3,2,1{S

)!(

!)1)...(2()1(),(

rn

nrnnnnrnP

n 1n 2n 1 rn

Example 2’. How many different ways are there to select 4 different players from 10 players on a team to play four tennis matches, where the matches are ordered ?Sol:

Example 4. 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:

.504078910)4,10( P

5040!7

Def. An r-combination of elements of a set is an unordered selection of r elements from the set.

Example 6 Let S be the set {1, 2, 3, 4}. Then {1, 3, 4} is a 3-combination from S.

Thm 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 :

)!(!!

!),()(),( rnr

nr

rnpnrrnCC n

r

nr 0

!),(),( rrnCrnP

稱為 binomial coefficient

Example 7. 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 1. Let n and r be nonnegative integers with r n. Then C(n,r) = C(n,nr)

pf : From Thm 2.

組合意義：選 r 個拿走 , 相當於是選 n r 個留下 .

),())!(()!(

!

)!(!

!),( rnnC

rnnrn

n

rnr

nrnC

Example 8. 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 11. 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, if there are 9 faculty members of the math department and 11 of the computer science department ?

Sol:

Exercise: 3, 11, 13, 21, 33, 34.

)4,11()3,9( CC

§4.5 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 =

∴

Thm 1. (The Binomial Theorem, 二項式定理 ) Let x,y be variables, and let n be a positive integer, Then

32233 ???))()(()( yxyyxxyxyxyxyx

)(32

)(32

333

232

231

330

3 )()()()()( yxyyxxyx

n

j

jjnnj

nnn

nnn

nnnnn yxyxyyxxyx0

11

110 )()()(...)()()(

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. (11)n = 0

25)32( yx 2525 ))3(2()32( yxyx

13122513 )3(2)(

)(...)()()()11( 210nn

nnnn

n

k

nk

k

0

0)()1(

n

k

nnn

nnnk

010 2)(...)()()(

Thm 2. (Pascal’s identity) Let n and k be positive integers with n k Then

k

n

k

n

k

n

1

1

PASCAL’s triangle

)(00

)(10 )(1

1

)(20 )(2

1 )(22

)(30 )(3

1 )(32 )(3

3

)(43

…

11 1

1 2 1

1 3 3 1

4

…

1 4 6 1

pf : ① 利用 來證

②( 組合意義 ):

)!(!

!),(

jij

ijiC

Suppose that T is a set containing n+1 elements. Let andA subset of T with k elements either containsa together with k1 elements of S,or contains elements of S. #

Ta }.{aTS

n

‧a

k

取法 =

a

k

n‧

0

a

k1

+n

‧

1

Thm 3. (Vandermode’s Identity)

pf :

r

k

knCkrmCrnmC

nmrrnm

0

),(),(),(

,0 ,,,

r

k

knCkrmC

rnmC

0

),(),(

個的方法 取)(

m n

m n m n m n↓↓ + ↓↓ +...+ ↓↓ 0, r 1, r1 r, 0

)!(!!

!),()(),( rnr

nr

rnpnrrnCC n

r

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

0 1 1 0 0 0 1 0

Each path can be Represented by a bit String consisting of m Os and n 1s.(→) (↑)

56

87 !3!5

!85

35

There are paths of the desired type.

n

nm

Exercise: 7, 21

§5.5 Generalized Permutations and Combinations

∴ 種方法 #

rnr 1

Thm 1. The number of r-permutations of a set of n objects with repetition allowed is n

r.

Thm 2. There are r-combinations from a set with n elements when repetition of elements is allowed.

pf : ( 視為有 r 個＊，要放入 n 個位置，即需插入 n1 個bar 將之隔開 )

),1( rnrC

＊＊｜＊｜｜＊＊＊a1 出現 2次

a2 出現 1 次 a3 不出現

a4 出現 3 次},,,,,{ 444211 aaaaaa

取法

},,,{ 4321 aaaa例： 設 n = 4, 集合為

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

have, where are nonnegative integers?’

Sol: 11 個 1 間要插入 2 個 bar

636

11321 xxx

321 ,, xxx

11211

種

※ 若上題改要求 則原式 可改成 此時相當於是 其中 且 ∴5 個 1 間要插入 2 個 bar 種 ( 注意： case

相當於 )

※ 若上題再改為 則需排除 ( 即 的情況 )

因 ∴ 共有 種

3 ,2 ,1 321 xxx

11321 xxx

532111)3()2()1( 321 xxx

5321 yyy ,2 ,1 2211 xyxy

321 ,, yyy

525

1y ,3 ,1 321 yy

4 x,5 ,2 321 xx

3 x,2 ,31 321 xx

31 x 41 x

2911)3()2()4( 321 xxx

222

525

※ Permutations with indistinguishable objects

Example 8. How many different strings can be made by reorderi

ng 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

)(73

)(42

)(21

)(11

))()()(( 11

21

42

73

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 :

※ Distributing objects into BoxesExample 9. 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

!!!

!

21 knnn

n

k

k

n

nnnn

n

nnn

n

nn

n

n 121

3

21

2

1

1

　　

!!!

!

21 knn

n

　　

547

552

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 .

Thm 4. The number of ways to distribute n distinguishable objects into k distinguishable boxes so that ni

objects are placed into box equals

( 跟 Thm 3 相等 )

Exercise: 15, 20, 17, 25, 31, 55

!!!

!

21 knnn

n

kii ,,2,1 ,

!32!5!5!5!5

!52

!32!5

!37

!37!5

!42

!42!5

!47

!47!5

!525

37542

547

552