r. johnsonbaugh, discrete mathematics 5th edition,...

Discrete Mathematics 6th edition, 2005

Chapter 6

Counting methods and the pigeonhole principle 1. Basic Principles 2. Permutations and Combinations 3. Algorithms for Generating Permutations and Combinations 4. Introduction to Discrete Probability* 5. Discrete Probability Theory* 6. Generalized Permutations and Combinations 7. Binomial Coefficient and Combinatorial Identities

2

6.1 Basic Principles

Menu for Quick Lunch 2 appetizers, 3 main courses,

4 beverages

24 possible dinners consisting of

1 appetizer, 1 main course and

1 beverage

NHT, NHM, NHC, NHR,

NCT, NCM, NCC, NCR,

NFT, NFM, NFC, NFR,

SHT, SHM, SHC, SHR,

SCT, SCM, SCC, SCR,

SFT, SNM, SFC, SFR

APPETIZERS

Nachos …………… 2.15 Salad ……………… 1.90 MAIN COURSES

Hamburger ……. 3.25 Cheeseburger... 3.65 Fish Filet ……….. 3.15 BEVERAGES

Tea ………………… .70 Milk ……………….. .85 Cola ………………. .75 Root Beer ……… .75

3

Multiplication Principle

Multiplication principle

If an activity can be performed in k successive

steps,

Step 1 can be done in n1 ways

Step 2 can be done in n2 ways

…

Step k can be done in nk ways

Then: the number of different ways that

the activity can be performed is the product

n1n2…nk

4

Addition Principle

How many 8-bit strings that begin either 101

or 111?

8-bit strings that begin 101

can be constructed in five successive steps

select the 4-th bit; select the 5-th bit;…

2 2 2 2 2 = 25 = 32 8-bit strings

8-bit strings that begin 111

can be constructed in five successive steps

select the 4-th bit; select the 5-th bit;…

2 2 2 2 2 = 25 = 32 8-bit strings

8-bit strings that begin either 101 or 111

32 + 32 = 64 8-bit strings

5

Addition Principle

Addition principle

Let X1, X2,…, Xk be a collection of k pairwise

disjoint sets, each of which has nj elements,

1 < j < k, then the union of those sets

k

X = Xj j =1

has n1 + n2 + … + nk elements

6

6.2 Permutations and Combinations

A permutation of n distinct elements x1, x2,…, xn

is an ordering of the n elements.

Example: there are 3! = 6 permutations of three

elements a, b, c:

abc bac cab

acb bca cba

Theorem 6.2.3

There are n! permutations of n elements

7

Theorem 6.2.10:

The number of r-permutations of a set of

n distinct objects is

P(n,r) = n(n-1)(n-2)…(n-r+1), rn

r-permutations

An r-permutation of n distinct elements is

an ordering of an r-element subset of

{x1, x2,…, xn}

Example: The number of 2-permutations of X={a,b,c} is 32 = 6

ab, ac, ba, bc, ca, cb

8

Combinations

Let X = {x1, x2,…, xn} be a set containing n distinct

elements

An r-combination of X is an unordered selection of r elements of X, for r < n

The number of r-combinations of n distinct element is denoted

C(n,r) or ( ) n r

Theorem 6.2.17:

The number of r-combinations of a set of n distinct

objects is

C(n,r) = = = , rn

P(n,r) n(n-1)(n-2)…(n-r+1) n! r! r! (n-r)!r!

9

6.6 Generalized Permutations and Combinations

Example

Problem

How many strings can be formed using the following letters?

M I S S I S S I P P I

Solution

11! ? no. why? duplication of letters!

Fill 11 blanks with the letters given . There are C(11,2) ways to choose positions for the two P’s.

There are C(9,4) ways to choose positions for the four S’s.

There are C(5,4) ways to choose positions for the four I’s.

There are 1 position to be filled by M.

By the multiplication principle, the number of ways of ordering the letters

C(11,2) C(9,4) C(5,4) = = = 34,650 11! 9! 5! 11!

2!9! 4!5! 4!1! 2!4!4!1!

2 Ps, 4 Ss, 4 Is and 1 M

10

Generalized Permutations and Combinations

Theorem 6.6.2:

Suppose that a sequence S of n items has n1 identical objects of type 1,

n2 identical objects of type 2, … and

nt identical objects of type t.

Then the number of orderings of S n!

n1!n2!...nt!

C(n, n1)C(n-n1, n2)C(n-n1-n2, n3) … C(n-n1-

…-nt-1, nt)

= …

=

n! (n-n1)! (n-n1- … -nt-1)!

n1!(n-n1)! n2!(n-n1-n2)! nt0!

n!

n1!n2!...nt!

Proof

11


Example

Problem Consider 3 books

a computer science book, a physics book, and a history book.

Suppose that the library has at least six copies of each of these books.

How many ways can we select six books?

Solution The problem is to choose unordered, repetitions allowed.

We use two bookends to separate 3 kinds of books.

By the bookends’ positions, the selection is determined. There are 8 positions: 6 books’ positions + 2 bookends’ positions

The possible number of bookends’ positions : C(8,2)

3 2 1 0 4 2 3 0 3

12


Theorem 6.6.5:

If X is a set containing t elements,

the number of unordered,

k-element selections from X,

repetitions allowed, is

C(k+t-1, t-1) = C(k+t-1, k)

Let X = {a1,…,at}

Consider k+t-1 slots and k+t-1 symbols (k ’s and (t-1) /’s)

A selection: each placement of these symbols into the slots

The number n1 of ’s up to the 1st /: the selection of n1 a1’s.

The number n2 of ’s between the 1st and 2nd /’s: the selection of n2 a2’s.

And so on.

There are C(k+t-1, t-1) way to select the positions for the /’s and

= The number of way to select the position for the ’s: C(k+t-1, k)

Proof

13

6.3 Algorithms for Generating Permutations and Combinations

Lexicographic order

Given two strings = s1s2…sp and = t1t2…tq

over {1,2,…,n}

Define < if

p < q and si = ti for all i = 1, 2,…, p

Or for some i, si ti and for the smallest i, si < ti

Example: if = 1324, = 1332, = 132,

then < and < .

14

Algorithm (1)

R-combination from set {x1, … ,xn}

String s1 … sr where s1 < s2 < s2 < s3 …

5 combination of { 1 2 3 4 5 6}

12345 ~ 23456

Find next r-combination, t, given a number, s

Find the right most element sm that is not at its max

ti = si for i = 1, .. , m -1

tm = sm + 1

tm+1 = sm + 2, … until tr

15

Pseudo code

Combination (r, n) { for i=1 to r, si = i; print(s1, …, sr); for i=2 to C(n,r) { m = r max_val = n while (sm == max_val) { m = m – 1; max_val = max_val – 1} sm = sm + 1; for j = m+1 to r sj = sj-1 + 1 print (s1, …, sr); } /* for loop */ }

1234 (first)

i = 2 to C(n,r) %exclude first one

m = 4, max_val = 6

skip while 1235

skip while 1236

sm = 6 (max)

while m = 3, max = 5

1245

m = 4 max = 6 again

skip while 1246

m = 4 max = 6


1256


1345

m = 4 max = 6 again

skip while

1346

…

16

Algorithm (2)

E.g. Permutation of {1 2 3 4 5 6} that follows 163542 Can it be 1635 _ _ ? No (only “4” and “2” left, “163524” is less)

Can it be 163 _ _ _ ? No (“4” “2” “5” left)

Remaining digits, 42 or 542, largest permutation

Must find the first digit d whose right neighbor r that satisfies d < r

d is the third digit 3 < 5

Digit following 16 must exceed 3 that is 4

164 _ _ _ 164 “2 3 5”

1234, 1243, 1324, 1342, 1423, … , 4231, 4312, 4321

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function perm (n, S) % print first perm disp(S); for (i = 2:factorial(n)), disp('=== Iteration:'); m = n - 1; while (S(m) > S(m+1)) m = m -1; end; disp('index of right most digit that is less than its right'); disp(m); k = n; while (S(m) > S(k)) k = k - 1; end; disp('index of right most digit that is greater than pivot'); disp(k);

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% swap sm and sk

temp = S(k);

S(k) = S(m);

S(m) = temp;

p = m+1;

q = n;

while (p<q)

temp = S(q);

S(q) = S(p);

S(p) = temp;

p = p + 1;

q = q - 1;

end;

disp(S);

end; % big for loop

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S = [5 6 7 8], n = 4

5 6 7 8 === Iteration: index of right most digit that is less than its right 3 index of right most digit that is greater than pivot 4 Swapping sm and sk 7 8 5 6 8 7

Who should change?

With who? Why is it this one?

20

5 6 8 7 === Iteration: index of right most digit that is less than its right: 2 index of right most digit that is greater than pivot: 4 Swapping sm and sk: 6 and 7 5 7 8 6 Sorting the rest 5 7 6 8 === Iteration: index of right most digit that is less than its right: 3 index of right most digit that is greater than pivot: 4 Swapping sm and sk: 6 and 8 5 7 8 6

7 is smallest that is greater than 6!

Until before 6, it is sorted from big to small

So once we make the switch start from beginning

And sort back to small to big

21

5 7 8 6 === Iteration: index of right most digit that is less than its right: 2 index of right most digit that is greater than pivot: 3 Swapping sm and sk: 7 and 8 5 8 7 6 Sorting the rest 5 8 6 7 === Iteration: index of right most digit that is less than its right: 3 index of right most digit that is greater than pivot: 4 Swapping sm and sk: 6 and 7 5 8 7 6

22

6.4 Introduction to Discrete Probability

An experiment is a process that yields an

outcome

An event is an outcome or a set of outcomes

from an experiment

The sample space is the event of all possible

outcomes

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Probability

Probability of an event is the number of outcomes in the event divided by the number of outcomes in the sample space.

If S is a finite sample space and E is an event (E is a subset of S) then the probability of E is

P(E) = |E| / |S|

Example 2 fair dice are rolled. What is the probability that

the sum of the numbers on the dice is 10?

There are 3 ways: (4,6), (5,5), (6,4)

The size of the event is 3

The probability is 3/36 = 1/12

24

6.5 Discrete Probability Theory

When all outcomes are equally likely and

there are n possible outcomes, each one

has a probability 1/n.

BUT this is not always the case. When

all probabilities are not equal, then some

probability (possibly different numbers)

must be assigned to each outcome.

25

Probability Function

A probability function P assigns to each outcome x

in a sample space S a number P(x)

so that 0P(x)1, for all xS, and

P(x) = 1 x S

Example 2 though 6 of die are equally likely appear

P(2) = P(3) = P(4) = P(5) = P(6)

1 is 3 times as likely as any other number to appear

P(1) = 3P(2)

1= P(1) + P(2) + P(3) + P(4) + P(5) + P(6)

= 3P(2) + P(2) + P(2) + P(2) + P(2) + P(2) = 8P(2)

P(2) = P(3) = P(4) = P(5) = P(6) = 1/8

P(1) = 3/8

26

Probability of an Event

The probability of an event E is defined as the

sum of he probabilities of the outcomes in E

P(E) = P(x) x E

Example probability of an odd number is

P(1)+P(3)+P(5) = 3/8+1/8+1/8 = 5/8

27


Suppose that E={x1, …, xk} and E={xk+1, …, xn}

Then

k n

P(E) = P(xi) and P(E) = P(xi) i=1 i=k+1

Now

k n P(E) + P(E) = P(xi) + P(xi)

i=1 i=k+1

n

= P(xi) = 1 i=1

Theorem 6.5.5

Let E be an event. The probability of E, the complement of E, satisfies

P(E) + P(E) = 1

Proof

28


Example: Birthday Problem

Find the probability that among n persons, at least two people have birthdays on the same month and date. Assume that all month and dates are equally likely, and

ignore February 29 birthdays.

E: the event “at least two persons have the same birthday”

E: the event “no two persons have the same birthday”

the size of the sample space: 365n

|E| = 365 364 … (365-n+1)

P(E) = [ 365 364 … (365-n+1) ] / 365n

P(E) = 1 - P(E)

n=22 P(E) = 0.475695, n=23 P(E) = 0.507297 > ½

29

Probability of Two Events

Theorem 6.5.9

Given any two events E1 and E2 in

a sample space S. Then

P(E1E2) = P(E1) + P(E2) – P(E1E2)

Let E1 = {x1, …, xi}

E2 = {y1, …, yj}

E1 E2 = {z1, …, zk}

Then in the list

x1, …, xi, y1, …, yj,

z1, …, zk occurs twice

Proof It follows that i j k P(E1E2) = P(xt) + P(yt) - P(zt)

t=1 t=1 t=1

= P(E1) + P(E2)

– P(E1E2)

30


Example

Among a group of students, some take art and some take computer science.

A student is selected at random.

Let A be the event “the student takes art,” and

let C be the event “the student takes computer

science.”

Then AC is the event “the student takes art

or computer science or both,” and

AC is the event “the student takes art and

computer science.”

31


Example

Two fair dice are rolled.

What is the probability of getting doubles or a sum of 6?

E1: get doubles

P(E1) = 6/36 = 1/6

E2: get a sum of 6

P(E2) = 5/36

E1 E2: get doubles and get a sum of 6

P(E1 E2) = 1/36

P(E1 E2) = 1/6 + 5/36 - 1/36 = 5/18

[(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)]

[(1,5), (2,4), (3,3), (4,2), (5,1)]

32

Mutually Exclusive Events

Events E1 and E2 are mutually exclusive

if and only if E1E2 = .

Example

E1: get doubles

P(E1) = 6/36 = 1/6

E2: get a sum of 5

P(E2) = 4/36 = 1/9

E1 E2: get doubles and get a sum of 5

P(E1 E2) = 1/6 + 1/9 = 5/18

Theorem 6.5.11

If E1 and E2 are mutually exclusive events,

P(E1 E2) = P(E1) + P(E2)

[(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)]

[(1,4), (2,3), (3,2), (4,1)]

33

Conditional Probability

Conditional probability is the probability of an event E,

given that another event F has occurred.

In symbols P(E|F).

If P(F) > 0 then

P(E|F) = P(EF) / P(F)

Example Weather records show that the probability of high barometric

pressure is 0.80, and

the probability or rain and high barometric pressure is 0.10.

The probability of rain given high barometric pressure is

P(R|H) = P(RH) / P(H) = 0.10 / 0.80 = 0.125

R: the event “rain”

H: the event “high barometric pressure”

34

Independent Events

If the probability of event E does not depend

on event F

P(E|F) = P(E)

E and F are independent events

P(E) = P(E|F) = P(EF) / P(F)

P(EF) = P(E)P(F)

Two events E and F are independent if

P(EF) = P(E)P(F)

35

Independent Events

Example

Intuitively, if we flip a fair coin twice, the outcome of

the second toss does not depend on the outcome of

the first toss

H: the event “head on first toss”

T: the event “tail on second toss”

HT: the event “head on first toss and

tail on second toss”

P(HT) = P(H)P(T) = ½ ½ = ¼

36

Pattern Recognition

Pattern recognition places items into classes,

based on various features of the items.

Given a set of features F we can calculate the

probability of a class C, given F: P(C|F)

Place the item into the most probable class, i.e.

the one C for which P(C|F) is the highest. Example: Wine can be classified as Premium (R), Table wine (T)

or Swill (S). Let F {acidity, body, color, price}

Suppose a wine has feature F, and P(T|F) = 0.5, P(R|F) = 0.2

and P(S|F) = 0.3. Since P(T|F) is the highest number, this wine

will be classified as table wine.

37

Bayes’ Theorem

Theorem 6.5.20

Given pairwise disjoint classes C1, C2,…, Cn and

a feature set F, then

P(Cj|F) = P(F|Cj)P(Cj)

n P(F|Ci)P(Ci)

i=1

P(Cj|F) = P(CjF)/P(F)

P(F|Cj) = P(FCj)/P(Cj)

P(Cj|F) = P(CjF)/P(F)

= P(F |Cj)P(Cj)/P(F)

Proof F = (FC1)(FC2)… (FCn)

Ci are pairwise mutually exclusive

(FCi) are pairwise mutually exclusive

P(F) = P(FC1)P(FC2)… P(FCn)

P(FCi) = P(F|Ci)P(Ci) n P(F) = P(F|Ci)P(Ci)

i=1

38

Example - Bayes’ Theorem (1/4)

The ELISA test is used to detect

antibodies in blood and indicates

the presence of the HIV virus.

15% of the patients at on clinic have

the HIV virus.

Among those that have the HIV virus,

95% test positive on the ELISA test.

Among those that do not have the

HIV virus, 2% test positive on the

ELISA test.

Problem: Find the probability that

a patient has the HIV virus if the

ELISA test is positive.

P(H)=0.15

P(Pos|H)=0.95 P(Pos|H)=0.02

P(H)=0.85

39


Problem: Find the probability that a patient has the HIV

virus if the ELISA test is positive.

P(H)=0.15

P(Pos|H)=0.95 P(Pos|H)=0.02

P(H)=0.85

A

B

P(H)=0.15

P(A)=0.15*0.95 P(B)=0.85*0.02

P(H)=0.85

A

B

40


Problem: Find the probability that a patient has the HIV

virus if the ELISA test is positive.

P(A) / P(A B)

P(A)=0.1425 P(B)=0.017

A

B

P(A | (A B)) = 0.1425 / (0.1425+0.017)

= 0.893

A

B

41


The ELISA test

is used to detect antibodies in blood and indicate the presence of the HIV virus.

15% of the patients at on clinic have the HIV virus.

Among those that have the HIV virus, 95% test positive on the ELISA test.

Among those that do not have the HIV virus, 2% test positive on the ELISA test.

Problem

Find the probability that a patient has the HIV virus if the ELISA test is positive.

Solution

H: the classes that are “has the HIV virus”

H: the classes that are “does not have the HIV virus”

Pos: the feature that is “test positive”

P(H)=0.15, P(H)=0.85, P(Pos|H)=0.95, P(Pos|H)=0.02

P(H|Pos) = { P(Pos|H) P(H) } / {P(Pos|H) P(H) + {P(Pos|H) P(H) }

= (0.95)(0.15) / {(0.95)(0.15) + (0.12)(0.85)} = 0.893

Bayes’ Theorem

42

Example: Telemarketing (1/4)

At the telemarketing firm SellPhone, Dale, Rusty, and Lee make calls.

What are P(D|H), P(R|H), P(L|H)?

Caller

Dale Rusty Lee

Percent of Calls 40 25 35

Percent of hang-ups 20 55 30

43


P(D) = 0.4

P(R) = 0.25

P(L) = 0.35

P(H|D) = 0.2 P(H|R) = 0.55

P(H|L) = 0.30

45


P(R|H) = = 0.426

P(L|H) = = 0.326

P(D|H) = P(H|D)P(D)

P(H)

P(R|H) = P(H|R)P(R)

P(H)

P(L|H) = P(H|L)P(L)

P(H)

0.08

0.3225 0.1375

0.3225

0.105

0.3225

P(D|H) + P(R|H) + P(L|H) = 1

P(D|H) = = 0.248

46

6.8 The Pigeonhole Principle

First form: If n pigeons fly into k pigeonholes and

k < n, some pigeonhole contains at least two

pigeons.

1 2 3

4 5 6

7 8 9

pigeonhole

1 2 3

4 5 6

7 8 9

1 2

3 4

5 6 7

8 9 10

11

12

47

Combinatorial Identities and Combinatorial Argument

Combinatorial Identity

An identity that results from some counting process

Combinatorial Argument

The argument that leads to its formulation

48

Second Form of the Pigeonhole Principle

If X and Y are finite sets with |X| > |Y| and

f : X Y is a function,

then f(x1) = f(x2) for some x1, x2 X, x1 x2.

a

b

c

d

e

f

1

2

3

4

5

X Y

49

Third Form of the Pigeonhole Principle

Let f be a function from a finite set X into a finite set Y.

Suppose |X| = n, |Y| = m.

Let k = n/m.

Then there are at least k values a1, a2,…, ak X

such that f(a1) = f(a2) = … f(ak).

Example:

n = 5, m = 3

k = n/m = 5/3 = 2.

a b c d e

1

2

3

X Y

50

Example 6.8.1

Ten persons have

first names Alice, Bernard, and Charles and

last names Lee, McDuff, and Ng.

Show that at least two persons have the same first and last

names.

There are nine possible names.

By the Pigeonhole principle, at least two persons have

the same last and first names.

Proof

51

Example 6.8.3

Show that if we select 151 distinct computer science courses numbered between 1 and 300 inclusive, at least two are consecutively numbered.

Let the selected course numbers be

C1, C2 , C3 , C4 , ... , C151

The 302 numbers consisting of the above together with

C1+1, C2 +1, , ... , C151+1

range in value between 1 and 301. By the second form of the

Pigeonhole Principle, at least two of these values coincide. It must

be that one of and one of are equal. Thus,

Ci = Cj +1, and course Ci follows course Cj

Proof

1

2

1 2

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