Discrete Mathematics

Chapter 5 Counting

大葉大學 資訊工程系 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 many possible passwords are there?

This section introducesa variety of other counting problemsthe basic techniques of counting.

§5.1 The Basics of counting

Ch5-3

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?

n1 n2

n1 + n2 ways

Ch5-4

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.

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

Ch5-5

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 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

Ch5-6

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

Ch5-7

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(a2), 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

#

Ch5-8

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

Ch5-9

Example 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 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 =26 ． 36 – 5 ∴26 + 26 ． 36 – 5 different names.

Ch5-10

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

Ch5-11

0

1

bit 1

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

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

Ch5-12

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=an1, a3=an2, …

1012 2

100...

98

100

99

100

100

100 )1( 100

Thm. 4 of §4.3 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

Ch5-13

§5.2 The Pigeonhole Principle (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 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.

Ch5-14

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 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

Ch5-15

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

912

100

Ch5-16

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

Exercise: 5, 13, 15, 31

Ch5-17

§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 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　　　□ □ □ …　　□　　　放法：

)!(

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

rn

nrnnnnrnP

n 1n 2n … 1 rn

Ch5-18

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:

9899100)3,100( P

5040!7

Ch5-19

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

)!(!!

!),()(),( rnr

nr

rnpnrrnCC n

r

nr 0

!),(),( rrnCrnP

稱為 binomial coefficient

Ch5-20

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,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

Ch5-21

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.

)4,11()3,9( CC

Ch5-22

§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

∴

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

)(32 )(3

2

333

232

231

330

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

n

j

jjnnj

nnn

nnn

nnnnn yxyxyyxxyx0

11

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

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

Ch5-23

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)(...)()()(

Ch5-24

Theorem 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

Ch5-25

pf : (algebraic proof, ① 代數證明 )

n

‧1

k

取法 =

1

k

n‧

0

1

k1

+n

‧

1

k

n

k

n

k

n

1

1

② (combinatorial proof, 組合意義證明 ):

)!1(!

)!1(1knk

nk

n

)!(!

!

)!1()!1(

!1 knk

n

knk

nkn

kn

)!1(!

!)1(

)!1(!

!)1(

)!1(!

!

knk

nn

knk

nkn

knk

nk

11

01

11

kn

kn

kn

Ch5-26

Theorem 3. (Vandermode’s Identity)

pf :

r

k

knCkrmCrnmC

nmrrnm

0

),(),(),(

,0 ,,,

r

k

knCkrmC0

),(),(

m n

r

C(m+n, r) = m n m n m n↓ ↓ + ↓ ↓ +...+ ↓ ↓ r, 0 r1, 1 0, r

=

r

nmnrmn

rm

0...110

Ch5-27

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

Red 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

!85

35

There are paths of the desired type.

n

nm

Exercise: 7, 21, 28