# §5.1 The Basics of counting

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## Mathematics

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### §5.1 The Basics of counting

Ú We will present two basic counting principles, the product rule and the sum rule.

Ú 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.

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

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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

Ν

x

x 100 1

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

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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

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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 n2 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 #

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Ú

-

-

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### Basic Counting Principles

Ú 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

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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.

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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

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Example 13

In 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 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.

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### ※ 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 0,1 0,1 → 共27

② . . . 0 0 → 共26

1 . . . 0 0 → 共25

∴ 27 +26 −25

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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 0 1

0 1

(0000) (0001) (0010) (0100) (0101) (1000) (1001) (1010)

∴ 8 bit strings 0

0 1 1

0

bit 3

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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=an−1, a3=an−2, …

101 2

2

... 100 98

100 99

100 100

100 )

1

( ⎟⎟ = = 100

⎜⎜ ⎞

⎝ + ⎛

⎟⎟ +

⎜⎜ ⎞

⎝ + ⎛

⎟⎟⎠

⎜⎜ ⎞

⎝ + ⎛

⎟⎟⎠

⎜⎜ ⎞

⎛ L

Thm. 4 of §5.3

1, 2 ,..., 100

) 2

( a a a

, ,...,

2 101

:

subset ∴ 100

不放

不放

不放

string.

⎢⎢

n

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### §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.

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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

1021 =3

⎥⎥

⎢⎢

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Example 5 Among 100 people there are at least

who were born in the same month. ( 100 objects, 12 boxes )

12 9 100 =⎥⎥

⎢⎢

Example 6 What is the minimum number of students required in a D.M class to be sure that at least six will

Sol: [N/5] = 6; N= 5*5+1 =26

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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

### sum ≤ 45

Some Elegant Applications of the Pigeonhole Principle

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Sol:

Let aj be 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

) 30 1

( ≤ j

30 2

1

### , a ,..., aa

j aj ≤ ∀

≤ 45 1

) 45 ...

1 ., .

(i e a1 < a2 < a3 < < a30

14 ,...,

14 ,

14 2 30

1 + a + a +

a

59 14

15 ≤ aj + ≤

) 59 14

...

14 14

14 15

., .

(i e a1 + < a2 + < a3 + < < a30 +

14 ,...,

14 ,

,..., 30 1 30

1 a a + a +

a j i and

., . (

14 i e j i

a

ai = j + +

#

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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)

9, 11, 4, 6 (8)

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

aN

a

a1, 2,...,

im

i

i

### a , ,...,

2 1

N i

i

i < < < m

...

1 1 2

)

., .

(i e 保持原順序

+1

i

i a

a

+1

i

i a

a

+1

< i

i a

a

+1

> i

i a

a

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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

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### §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

□ □ □ … □

↑ ↑ ↑ … ↑ 放法：

### = { S

)!

( ) ! 1 )...(

2 (

) 1 (

) ,

( n r

r n n n

n n r

n

P = ⋅ − ⋅ − − + = −

n−1

### n − 2

n−r +1

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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:

### 7 =

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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 :

)!

(

!

!

! ) ,

(

nr p rn r r nn r

rn

n r

≤ 0

### P = ×

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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,n−r) pf : From Thm 2.

) ,

))! ( (

( )!

(

! )!

(

! ) !

,

( C n n r

r n n

r n

n r

n r r n

n

C =

=

=

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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.

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### §4.5 Binomial Coefficients (二項式係數)

Example 1.

∴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

3 2

2 3

3 ( )( )( ) ? ? ?

)

(x + y = x + y x + y x + y = x + x y + xy + y

) (32

) (32

3 3

3 2

3 2 2

3 1 3

3 0

3

### ∑

=

+ + + =

+

=

+ n

j

j j n n

j n

n n n

n n n

n n

n

n x x y xy y x y

y x

0 1

1 1

1

0) ( ) ... ( ) ( ) ( )

( )

(

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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

3 2

( x y

25

25 (2 ( 3 ))

) 3 2

( x y = x + y

13 12

25

13 ) 2 ( 3)

(

n

0n

1n

2n

nn

=

=

n

k

n k k 0

0 )

( ) 1 (

### ∑

=

= +

+ +

n =

k

n n

n n

n n

k 0

1

0) ( ) ... ( ) 2 (

) (

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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 (11)

)

( 02 (12 ) (22 )

)

(30 (13 ) (32 ) (33 )

) (34

1

1 1

1 2 1

1 3 3 1

4

1 4 6 1

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pf : ①利用 來證

②(組合意義):

)!

(

! ) !

,

( j i j

j i i

C = −

Suppose that T is a set containing n+1 elements.

Let and

A subset of T with k elements either contains a together with k1 elements of S,

or contains elements of S. # T

aS = T{a}.

n

a

k

=

a

k n

‧ 0

a

k−1 +

n

‧ 1

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Thm 3. (Vandermode’s Identity)

pf :

= +

= +

≤ Ζ

r

k

k n C k

r m C r

n m

C

n m r

r n m

0

) , ( )

, ( )

, (

, 0

, ,

,

### ∑

=

=

=

= +

r

k

k n C k

r m C r

n m

C

0

) , ( )

, (

) (

m n

m n m n m n

↓↓ + ↓↓ +...+ ↓↓

0, r 1, r−1 r, 0

)!

(

!

!

! ) ,

) (

( ) ,

(n r nr p rn r r nn r C

Crn = = = =

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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 8 7

! 3

! 5

! 8 5

3 5

=

×

=

=

⎛ +

There are paths of the desired type.⎟⎟

⎜⎜

⎛ + n

n m

Exercise: 7, 21

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### §5.5 Generalized Permutations and Combinations

+ 種方法 # r

n

r 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 is allowed.

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 (r n r C + −

＊＊｜＊｜｜＊＊＊

a1出現2次

a2出現1次 a3不出現

a4出現3次

} ,

, , , ,

{a1 a1 a2 a4 a4 a4

} ,

, ,

{a1 a2 a3 a4 例： 設n = 4, 集合為

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Example 4. Suppose that a cookie shop has 4 different kinds of cookies. How many different ways can

Example 5. How many solutions does the equation have, where are nonnegative integers?’

Sol: 11個1間要插入2個bar

⎛ + 6

3 6

3 11

2

1 + x + x = x

3 2 1, x , x x

⎛ + 11

2

11

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※若上題改要求 則原式

∴5個1間要插入2個bar 種 (注意： case

※若上題再改為

∴共有

3

, 2

,

1 2 3

1 x x

x

3 11

2

1 + x + x = x

5 3 2 1 11 )

3 (

) 2 (

) 1

(x1 + x2 + x3 = =

3 5

2

1 + y + y =

y y1 = x1 1, y2 = x2 2, Ν

3

2 1, y , y y

⎛ + 5

2 5

1 y , 3

,

1 2 3

1 = y = =

y

4 x

, 5

,

2 2 3

1 = x = =

x

3 x

, 2

, 3

1 x1 x2 3

1 > 3

x x1 > 4

2 9 11 )

3 (

) 2 (

) 4

(x1 + x2 + x3 = =

⎛ +

⎛ +

2 2 2 5

2 5

(35)

※ Permutations with indistinguishable objects

Example 8. How many different strings can be made by reordering the letters of the word

SUCCESS ? Sol:

∴共

) (73

) (42

) (12

) (11

) )(

)(

)(

(73 42 12 11

(36)

Thm 3. The number of different permutations of n objects, where type 1： n1

type 2： n2 is type k： nk

pf :

※ Distributing objects into Boxes

Example 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：

player 2： 從剩下的牌再發5張

!

!

!

!

2

1 n nk

n

n

M L

⎟⎟

⎜⎜

⎟⎟

⎜⎜

⎟⎟

⎜⎜

⎟⎟⎛ −

⎜⎜

k

k

n

n n

n n n

n n n n

n n n

n 1 2 1

3 2 1 2

1 1

L L

!

!

!

!

2

1 n k

n n

L

=

5 47

5 52

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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

!

!

!

!

2

1 n nk

n

n L

k i

i, =1,2,L,

! 32

! 5

! 5

! 5

! 5

! 52

! 32

! 5

! 37

! 37

! 5

! 42

! 42

! 5

! 47

! 47

! 5

! 52 5

37 5

42 5

47 5

52 = =

Updating...

## References

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