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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)

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discrete mathematics

Discrete Mathematics

Chapter 5 Counting

大葉大學 資訊工程系 黃鈴玲

slide2

§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.
basic counting principles

n1

n2

n1 + n2 ways

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+n2ways 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?

slide4

n1

n1 × n2

ways

n2

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

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.

slide5

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

Example 4 How many different bit strings are there of length seven?

Sol:123456 7

□ □ □ □ □ □ □

↑ ↑ ↑ ↑ ↑ ↑ ↑

0,1 0,1 0,1 . . . . . . 0,1

→ 27種

slide6

a1

a2

am

b1

b2

bn

f

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

slide7

Example 7 How many one-to-one functions are there

from a set with melements 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 #

slide8

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

slide9

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 Vibe the number of variable names of length i.

V1 =26

V2 =26.36 – 5

∴26 + 26.36 – 5 different names.

slide10

※ The Inclusion-Exclusion Principle (排容原理)

A B

Example 17How 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 10,1 0,1 → 共27種

②  . . . . . . . . . . . . 0 0 → 共26種

③ 1 . . . . . . . . . . . 0 0 → 共25種

 27 +26 -25 種

slide11

(0000)

0

0

1

(0001)

0

(0010)

0

∴ 8 bit strings

0

1

(0100)

0

1

0

(0101)

1

0

0

(1000)

0

1

1

(1001)

1

(1010)

0

bit 1

bit 3

※ Tree Diagrams

Example 18 How many bit strings of length four do not have two consecutive 1s ?

Sol:

Exercise: 11, 17, 23, 27, 38, 39, 47, 53

slide12

Ex 38. How many subsets of a set with 100 elements

have more than one element ?

Sol:

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 a1=an, a2=an-1,a3=an-2, …

Thm. 4 of §4.3

空集合及只有1個元素的集合

不放

不放

不放

slide13

§5.2 The Pigeonhole Principle (鴿籠原理)

Theorem 1 (The Pigeonhole Principle)

Ifk+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 thek 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.

slide14

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 Nobjects 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.

slide15

Example 5 Among 100 people there are at least

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

slide16

(跳過)

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.

存在一段時間的game數和=14

slide17

(跳過)

Sol:

Let ajbe 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

slide18

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 calleddecreasing(遞減) if

A sequence is called strictly increasing(嚴格遞增)if

A sequence is calledstrictly decreasing(嚴格遞減)if

slide19

(跳過)

Theorem 3. Every sequence of n2+1distinct 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, 15, 31

slide20

§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 anr-permutation.

Example 2.LetS ={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 withn

distinct elements is

位置: 1 2 3 … r

   □ □ □ … □

放法:

slide21

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:

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:

slide22

稱為 binomial coefficient

Def. An r-combination of elements of a set is an

unordered selection of r elements from the set.

Example 9Let 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 :

slide23

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.

組合意義:選 r 個拿走,相當於是選 n - r 個留下.

slide24

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:

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

slide25

§5.4 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 =

Theorem 1.(The Binomial Theorem, 二項式定理)

Let x,y be variables, and let n be a positive integer, then

slide26

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

slide27

Theorem 2.(Pascal’s identity)

Let nand k be positive integers with n k

Then

PASCAL’s triangle

slide28

n

1

n

n

1

1

+

k

k

0

k-1

1

pf : ①(algebraic proof, 代數證明)

②(combinatorial proof, 組合意義證明):

取法=

slide29

m

n

r

Theorem 3.(Vandermode’s Identity)

pf :

C(m+n, r) =

m n m n m n

↓ ↓ + ↓ ↓ +...+ ↓ ↓

=

r,0 r-1, 0 0, r

slide30

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

Each path can be represented by a bit string consisting of m0s and n1s.

(→) (↑)

Red path對應的字串: 0 1 1 0 0 0 1 0

There are paths of the desired type.

Exercise: 7, 21, 28