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 arethere 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 integernot exceeding 100. What is the largest number of chairs that can be labeled differently?

Sol: 26 × 100 = 2600 ways to label chairs.letter

Ν∈≤≤

xx 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 5How many different license plates (車牌) are

available if each plate contains a sequence of 3 lettersfollowed by 3 digits ?

Sol: □ □ □ □ □ □ →263．103

letter digit

Example 6 How many functions are there from aset 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

a1a2．．．

am

b1b2．．．bn

f

Example 7 How many one-to-one functions are therefrom 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 #

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 PrinciplesThe 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 11Suppose 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=6,7,8P6 = 366 − 266P7 = 367 − 267P8 = 368 − 268

∴ P6 + P7 + P8 = 366 + 367 + 368 − 266 − 267 − 268種

Example 13In a version of Basic, the name of a variable

is 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 stringsof 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 =26V2 =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 IU −+=

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

00

0

01

1

10

01

(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 elementshave more than one element ?

Sol:

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

Sol: If a1a2 ... an is a palindrome, then a1=an, a2=an−1, a3=an−2, …

1012 2

100...

98100

99100

100100

)1( 100 −==⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛L

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 wouldbe at most k. This is a contradiction.

Example 1. Among any 367 people, there must beat least two with the same birthday, because thereare 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 toguarantee that at least two students receive thesame 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, thenthere is at least one box containing at leastobjects.

e.g. 21 objects, 10 boxes ⇒ there must be one box containing at least objects.

⎥⎥⎤

⎢⎢⎡

kN

31021

=⎥⎥⎤

⎢⎢⎡

Example 5 Among 100 people there are at leastwho 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 baseballteam plays at least 1 game a day, but no morethan 45 games. Show that there must be a periodof some number of consecutive days during which theteam 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 distinctintegers 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 ≤

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, 7subsequence: 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

miiiaaa ,...,,

21Niii m ≤

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 relements of a set is called an r-permutation.

Example 1. Let The arrangement 3,1,2 isa permutation of S. The arrangement 3,2 is a2-permutation of S.

Thm 1. The number of r-permutation of a set with ndistinct elements is

位置： 1 2 3 … r□ □ □ … □↑ ↑ ↑ … ↑

放法：

}.3,2,1{=S

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

nrnnnnrnP−

=+−−⋅−⋅=

n 1−n 2−n 1+−rn

Example 2’. How many different ways are there to select4 different players from 10 players on a team to playfour tennis matches, where the matches are ordered ?

Sol:

Example 4. Suppose that a saleswoman has to visit8 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 anunordered 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 withn elements, where n is a positive integer and r isan integer with , equals

pf :

)!(!!

!),()(),( rnr

nr

rnpnrrnCC

nr −====

nr ≤≤0

!),(),( rrnCrnP ×=

稱為 binomial coefficient

Example 7. We see that C(4,2)=6, since the2-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,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.

),())!(()!(

!)!(!

!),( rnnCrnnrn

nrnr

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 3faculty members from the math department and 4from the computer science department, if there are9 faculty members of the math department and11 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. (1−1)n = 0

25)32( yx −2525 ))3(2()32( yxyx −+=−

13122513 )3(2)( −⋅⋅

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

nnnn ++++=+

∑=

=−n

k

nk

k

00)()1(

∑=

=+++=n

k

nnn

nnnk

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

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

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +kn

kn

kn

11

PASCAL’s triangle)( 00

)(10 )(11

)( 20 )(21 )(

22

)( 30 )(31 )(

32 )(

33

)( 43

…

11 1

1 2 11 3 3 1

4

…

1 4 6 1

pf : ①利用 來證

②(組合意義):

)!(!!),(

jijijiC−

=

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

Ta∈ }.{aTS −=

n‧

a

k

取法=

a

k

n‧

0

a

k−1

+n

‧

1

Thm 3. (Vandermode’s Identity)

pf :

∑=

+

⋅−=+

≤≤Ζ∈r

kknCkrmCrnmC

nmrrnm

0),(),(),(

,0 ,,,

∑=

⋅−=

==+

r

kknCkrmC

rnmC

0),(),(

個的方法 取)(

m n

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

)!(!!

!),()(),( rnr

nr

rnpnrrnCC

nr −====

Ex 33. Here we will count the number of pathsbetween the origin (0,0) and point (m,n) suchthat each path is made up of a series ofsteps, where each step is a move one unitto 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 10 1 1 0 0 0 1 0

Each path can beRepresented by a bitString consisting of mOs 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 nr.

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

pf : (視為有 r 個＊，要放入 n 個位置，即需插入 n−1 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 differentkinds of cookies. How many different ways can6 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 =++ xxx321 ,, 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 ≥≥≥ xxx11321 =++ xxx

532111)3()2()1( 321 =−−−=−+−+− xxx

5321 =++ yyy …−=−= ,2 ,1 2211 xyxyΝ∈321 ,, yyy

⎟⎠⎞⎜

⎝⎛ +

525

1y ,3 ,1 321 === yy4 x,5 ,2 321 === xx

3 x,2 ,31 321 ≥≥≤≤ xx31 >x 41 >x

2911)3()2()4( 321 =−=−+−+− xxx⎟⎠⎞⎜

⎝⎛ +−⎟

⎠⎞⎜

⎝⎛ +

222

525

※ Permutations with indistinguishable objectsExample 8. 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

)(73)(42)(21)(11

))()()(( 1121

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 knnnnLM

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−−−⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛ −k

k

nnnnn

nnnn

nnn

nn 121

3

21

2

1

1

　　LL

!!!!

21 knnn　L

=

⎟⎠⎞⎜

⎝⎛

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 2box 3 的給 player 3box 4 的給 player 4

而 box 5 的是剩下的牌.

Thm 4. The number of ways to distribute n distinguishableobjects into k distinguishable boxes so that niobjects are placed into box equals

(跟Thm 3相等)

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

!!!!

21 knnnnL

kii ,,2,1 , L=

!32!5!5!5!5!52

!32!5!37

!37!5!42

!42!5!47

!47!5!52

537

542

547

552 =⋅⋅⋅=⎟

⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛∴

Discrete Mathematics §5.1 The Basics of counting Basic Counting Principles