Discrete Structure Chapter 4 Counting 52

8/10/2019 Discrete Structure Chapter 4 Counting 52

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Chapter 4

Counting


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4.1 Rule of Sum

4.2 Rule of Product

4.3 Permutations

4.4 Combinations

4.5 Summation Notation

4.6 Combinations with Repetition

4.7 Summary

Counting


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Counting

4.1: The Rule of Sum

The Rule of Sum:

m ways n ways

cannot be done simultaneously

then performing eithertask can be accomplished in any one of

m+n ways


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Counting

4.1: The Rules of Sum

Eg: 40 textbooks on sociology

50 textbooks on anthropology

to select 1 book: 40+50 choices


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Eg:A Computer Science instructor has seven introductory

books of each on C++, Java and Perl

Soln:

How many ways can a student select a book?

7 + 7 + 7 = 21 ways

Counting


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Counting


Eg: things 1 2 3 ... kways m1 m2 m3 mk

select one of them: m1 +m2 +m3+...+ mk ways


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4.2: The Rules Product

Eg: An administrator assigns 12 of her employees to twocommittees.

Soln:Committee A - 5 members

Committee B7 members

Therefore, he can select members for committee A and B in5 x 7 = 35 ways

Counting


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Counting


The Rule of Product

m ways n ways

then performing this task can be accomplished in any one ofmn ways


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Eg:A drama club of Central University is holding an auditionto select a hero and a heroin for a play.

If six men and eight women came for the audition, how manyways can we select a hero and a heroin?

Soln:

6 x 8 = 48 ways

Counting



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Counting

4.2: The Rules of Product

Eg: The license plate: 2 letters-4 digits

(a) no letter or digit can be repeated26 25 10 9 8 7

(b) with repetitions allowed

26 26 10 10 10 10

(c) same as (b), but only vowels and even digits

52 x 54

= 3,276,000 different possible plates

= 6,760,000 different possible plates


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Eg:Mrs. Foster operates a Quick Snack Coffee Shop at AWL corporation.The menu at her shop is limited: six kinds of muffins, eight kinds ofsandwiches and five beverages(hot coffee, hot tea, iced tea, cola andorange juice).

Ms Dodd, an editor at AWL, sends her assistant Carl to the shop to gether lunch either a muffin and a hot beverage or a sandwich and a coldbeverage.

Soln:

A muffin and a hot beverage = 6 x 2 = 12 ways

A sandwich and a cold beverage = 8 x 3 = 24 ways

There are 12 + 24 ways to buy Ms Dodds lunch

Counting

4.1 and 4.2: The Rules of Sum and Product


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Counting


BASIC variables: single letter

or single letter and single digit

26+26x10=286

rule of sum rule of product


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Eg:Counting Passwords with Three or Fewer Letters

A computer access password consists of from one to three

letters chosen from the 26 in the alphabet with repetitions

allowed. How many different passwords are possible?

Soln:

Counting



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Eg:Counting the Number of Integers Divisible by 5

How many three digit integers(integers from 100 to 999 inclusive) are

divisible by 5?

Soln:

Counting

4.2: The Rules of Sum and Product


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Eg: Number of Personal Identification Numbers(PINS)

A typical PIN(Personal Identification Number) is a sequence of any four

symbols chosen from the 26 letters in the alphabet and the ten digits,

with repetition allowed. How many different PINS are possible?

Soln:

Counting



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Eg: The number of Elements in a Cartesian Product

Suppose A1, A2, A3and A4 are sets with n1, n2, n3and n4

elements, respectively. Show that the set

A1x A2x A3x A4 has n1n2 n3n4 elements.

Soln:

Counting



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Eg:Number of PINS without repetition

A typical PIN(Personal Identification Number) is a sequence of any

four symbols chosen from the 26 letters in the alphabet and the ten

digits, with repetition is not allowed.. How many different PINS are

possible?

Soln:

Counting



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Eg: Counting the following nested loop:

for I:= 1 to 4

for j:= 1 to 3

[Statements in body of inner loop. None contain branching statements

that lead out of the inner loop.]

next jnext i

How many times will the inner loop be iterated when the algorithm is

implemented and run?

Soln:

Counting



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When the Product Rule is Difficult or Impossible to Apply:

Consider the following problem:

Three officersa president, a treasurer and a secretary -- are to be chosen from amongfour people: Ann, Bob, Cyd and Dan.

Suppose that, for various reasons, Ann cannot be president and either Cyd or Dan must besecretary. How many ways can the officers be chosen?

By Product Rule, a person might answer as follows:

3 choices for president(all except Ann)

3 choices for treasurer(all except the one chosen as president)2 choices for secretary (Cyd or Dan)

Therefore, by product rule, 3 x 3 x 2 = 18 choices in all.

Unfortunately, this analysis is incorrect.

Why?

The no. of ways to choose the secretary varies depending on who is chosen for presidentand treasurer. For instance, if Bob is chosen for president and Ann for treasurer, then thereare two choices for secretary (Cyd and Dan). But if Bob is chosen for president and Cyd fortreasurer, then there is just

one choice for secretary.

The clearest way construct a tree diagram.

Counting


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Counting

4.3: Permutations

Note: For an integer n, nfactorial (denoted n!) is defined by

0!=1, (n = 0)

n!=(n)(n-1)(n-2)...(3)(2)(1), for n>1.

Def:A permutation of a set of distinct objects is an ordered arrangement of

these objects.

Eg:Consider the set of elements a, b , c

abc acb cba bac bca cab

The number of permutations of three elements are

3! = 3 x 2 x 1 = 6 permutations


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Counting

4.3: Permutations

n n n n r n

n rP n r

( ) ( ) . . . ( )

!

( ) !( , )1 2 1

if repetitions are allowed:nr

In general:

Given a set of n distinct objects, how many permutations does the set have?

Step 1: Choose 1st element ( n ways to perform this)

Step 2: Choose 2ndelement ( n1 ways to perform this)

Step n: Choose n th element (1 way to perform this)

Therefore, for n objects n!

The number of r-permutation of a set with n-elements is denoted by P(n,r) ,

where (1 < r < n ).


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Proof:

Step 1: Choose 1stelement ( n ways to perform this)

Step 2: Choose 2nd

element ( n

1 ways to perform this)

Step 3: Choose 3rdelement (n 2ways to perform this)

Step r: Choose r th element (n - ( r - 1) way to perform this)

Therefore, by Rule of Product:

nx (n 1)x (n 2)x x (n r + 1 )

Counting4.3: Permutations

n n n n r n

n r

P n r

( ) ( ) . . . ( ) !

( ) !

( , )1 2 1


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Eg: Given the set {a , b , c }

ab ac ba bc ca cb

This is called 2-permutation of {a , b , c }

Counting

4.3: Permutations


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Counting

4.3: Permutations

E.g: a) permutation of letters BALL

4!/2!=12

b) permutation of PEPPER6!/(3!2!)=60

c) permutation of MASSASAUGA

10!/(4!3!)=25200

if all 4 As are together

7!/3!=840


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Eg:a) How many 4-permutations are there of a set of seven objects?b) How many 5-permutations are there of a set of five objects?

Soln:

Counting

4.3: Permutations


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Eg:a) How many different ways can three of the letters of the word

BYTES be chosen and written in a row?

b) How many different ways can this be done if the first letter must be B?

Soln:

Counting

4.3: Permutations


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Counting

4.3: Permutation

Eg: Permutations of the Letters in a word

a) How many ways can the letters in the word COMPUTER

be arranged in a row?

b) How many ways can the letters in the word COMPUTER

be arranged if the letters CO must remain next to eachother(in order) as a unit?


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Counting4.3: Permutations

Eg: Number of Manhattan paths between

two points (2,1) and (7,4) with integer coordinated

Soln:

From (2,1) to (7,4): 3 Ups, 5 Rights

Each permutation of UUURRRRR is a path.

8!/(5!3!)=56

y y

x x0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

(2,1)

(7,4)

(2,1)

(7,4)


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Eg: Proving a property of P( n , r )

Prove that for all integers n > 2 ,

P( n , 2 ) + P( n , 1 ) = n2

Soln:

Counting

4.3: Permutations


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Counting

4.4: Combinations(The Binomial Theorem)

When dealing with any counting problem, we should ask

ourselves about the importance of order in the problem.

When order is relevant, we think in terms of permutations and

arrangements and the rule of product. When order is not

relevant, combinations could play a key role in solving the

problem.


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C i


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eg.:Let S = {Ann, Bob, Chan, Dan}. Each committee consisting of three of the four

people in S is a 3-combination of S.

List all such 3-combination of S.

Solution:

{Bob, Chan, Dan}{Ann, Chan, Dan}

{Ann, Bob, Dan}

{Ann, Bob, Chan}

Counting


C ti


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Eg:Unordered Selection

a) How many ways are there to select five players from a 10 member tennis

team to make a trip to a match at another school?

b) How many unordered selections of two elements can be made from the

set {0 , 1 , 2 , 3}?

Soln:

Counting


a) C(10,5) =252

b) C(4,2)=6{0,1},{0,2},{0,3},{1,2},{1,3},{2,3}

C ti


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Eg:Relation Between Permutations & CombinationsWrite all 2-permutation

from the set {0 , 1 , 2 , 3}? Find P(4,2).

a) Write all 2-combination from the set {0 , 1 , 2 , 3}? Find C(4,2).

Soln:

Counting


C ti


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Counting


Select 3 cards from a deck of playing cards without replacement:

order of selection is relevant: P(52,3)= 52 51 50

order of selection is irrelevant: P(52,3)/3!=C(52,3)

Eg:A standard deck of playing cards consists of 52 cards comprising four suits:

Clubs, Diamonds, Hearts, Spades.

Each suit has 13 cards: ace, 2 , 3 , 4 ,., 9 , 10 , jack , queen , king.


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Counting


Eg:At Sultan Ismail High School, the PE teacher must select nine girls fromthe junior and senior classes for a volleyball team.

a) If there are 28 juniors and 25 seniors, how many selection can she make?

b) If two juniors and one senior are the best spikers and must be on the

team, the how many way to select the team?

c) For a certain tournament the team must comprise four juniors and fiveseniors. How many ways to select the team?

a) C(53,9) = 4,431,613,550 ways

b) C(50,6) = 15,890,700 ways

c)By Rule of Product: C(28,4) x C(25,5) = 1,087,836,750 ways


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Eg:A PE teacher must make up four basketball teams of nine girls each from 36

freshman girls in her P.E. class. In how many ways can she select these four

teams?

Counting


C(36,9) x C(27,9) x C(18,9) x C(9,9) = 2.145 x 10 19


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Counting

Eg:The letters in TALLAHASSEE

a) How many arrangements can be made from the letters in TALLAHASSEE?

b) How many of these arrangements have no adjacent As ?

11

3 2 2 2 1 1831600

!

! ! ! ! ! !,

b) without adjacent A:

82 2 2

93

!! ! !

disregard A first 9 positions for 3 A

to be inserted


E E S T L L S H

Counting


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Counting4.4: Combinations(The Binomial Theorem)

Eg: How many ways are there to select a committee members to develop a

discrete mathematics course at a school if the committee is to consist of three

faculty members from the mathematics department and four from the computer

science department, if there are nine faculty members of the mathematics

department and eleven members of the computer science department?

Soln:

Counting


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Eg:a) A student taking a history examination is directed to answer any seven of 10 essay

questions. There is no concern about order here, so how many ways can the student

answer the examination?

b) If the student must answer three questions from the first five and four questions from

the last five, how many ways the student can complete the examination?

c) Finally, should the direction on this examination indicate that the student must answer

seven of the 10 questions where at least three are selected from the first five, how

many ways can the student complete the examination?Soln:

Counting4.4: Combinations(The Binomial Theorem)

Counting


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Counting4.6: Combinations with Repetition

Eg: On their way home from track practice, seven high school freshmen stop at a

restaurant. Each of them bought one of the following: a cheeseburger, a hot dog, a

taco, or a fish sandwich. How many different purchases are possible?

1. c , c , h , h , t , t , f

2. c , c , c , c , h , t , f

3. c , c , c , c , c , c , f

4. h , t , t , f , f , f , f

5. t , t , t , t , t , f , f

6. t , t , t , t , t , t , t

7. f , f , f , f , f , f ,f

1. x x |x x | x x | x

2. x x x x | x | x | x

3. x x x x x x | | | x

4. | x | x x | x x x x

5. | | x x x x x | x x

6. | | x x x x x x x |

7. | | | x x x x x x x

Note:In general, the number of r-combinations from a set with n elements when

repetition of elements is allowed is given by C ( n + r - 1 , r )

No. of bars = n1 ; no. of star = r (i.e. 7-combinations of 4 elements)

The number of combinations = C ( n + r - 1 , r ) = C ( 4 + 71, 7) = C (10 ,7 )


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Counting

4.5: Summation Notation

Counting


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Eg:A donutshop offers 20 kinds of donuts. Assuming that there are at least adozen of each kind when we enter the shop. How many ways can we select

a dozen of donuts?

Soln: ( 12-combinations of 20 elements)

C ( n + r - 1 , r ) = C ( 20 + 121, 12) = C (31 , 12) = 141,120,525 ways.

Counting

4.6: Combinations with Repetition

Counting


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Eg: In how many ways can we distribute seven bananas and six oranges amongfour children so that each child receives at least one banana?

Soln:

After giving one child a banana, the remaining three bananas are distributed to

four children in:

C (4 + 31, 3) = 20 ways.

Six oranges are distributed in C (4 + 61, 6) = 84 ways.

Therefore, to distribute seven bananas and six oranges = 20 x 84 = 1680 ways.

Counting


Counting


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Counting


Eg:How many ways are there to select five bills from a cash box containing $1

bills, $2 bills, $5 bills, $10 bills, $20 bills, $50 bills and $100 bills?

Assume that the order in which the bills are chosen does not matter, that the bills

of each denomination are indistinguishable, and that there are at least five bills of

each type.

Soln:

Examples of ways to select five bills:

| | | * * | | | * * *

Or * | * | * * | | * | |

Or * | | | * * | | * | *

No. of bars = n1 ; no. of stars = r (i.e. 5-combinations of 7 elements)

C ( n + r - 1 , r ) = C (7 + 51, 5) = C (11, 5) = 462 ways.

$100 $50 $20 $10 $5 $2 $1

Counting


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Eg: Counting Iterations of a Loop

How many times will the innermost loop be iterated when the algorithm segment below is

implemented and run? (Assume n is a positive integer.)

for k:= 1 to n

for j:= 1 to k

for i:=1 to j

[Statements in the body of the inner loop, none containing

branching statements that lead outside the loop.]next I

next j

next k

Soln:

Counting


K 1 2 3 .. n

J 1 1 2 1 2 3 .. 1 2 .. n

i 1 1 1 2 1 1 2 1 2 3 1 1 2 .. 1 .. N

The number of iteration of the innermost loop= [n(n + 1)(n + 2)]/6

Counting


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Exercise:How many ways are there to place ten indistinguishable balls

into eight distinguishable bins?

Soln:

Counting


Counting


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Exercise:Suppose that a cookie shop has four different kinds of cookies.

How many different ways can six cookies be chosen?

Assume that only the type of cookie, and not the individual cookies or

the order in which they are chosen, matters.

Soln:

Counting



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Counting

4.7 Summary

order is repetitions

relevant are allowed type of result formula

YES NO permutation

YES YES arrangement

NO NO combination

combination

NO YES with repetition

P n r n n r

r n

( , ) !/ ( ) !,

0

n n rr

, , 0

C n r n r n r

r n

( , ) !/ [ !( ) !]

0

n r

r

1

Select or Order robjects from ndistinct objects


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How do I know what to multiply and what to add?

When to use rule of product and when to use rule of sum?

Tips:

1. You need to imagine the objects you are to count.

2. Make an actual list of the items you are trying to count to get a sense

for how to obtain them in a systematic way.

3. Construct a model that would allow you to continue counting the

objects.

Counting

4.7 Summary


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Counting

End of Chapter 4

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Discrete Structure Chapter 4 Counting 52