Counting Fundamentals

Counting FundamentalsCounting Fundamentals

Ginger Holmes Rowell, Ginger Holmes Rowell, Middle TN State UniversityMiddle TN State University

MSP WorkshopMSP WorkshopJune 2006June 2006

ObjectivesObjectives

• Solve basic problems that involve Solve basic problems that involve counting techniques, including the counting techniques, including the multiplication principle, permutations, multiplication principle, permutations, and combinations; use counting and combinations; use counting techniques to understand various techniques to understand various situations (e.g., number of ways to situations (e.g., number of ways to order a set of objects, to choose a order a set of objects, to choose a subcommittee from a committee, to subcommittee from a committee, to visit visit n n cities)cities)

OverviewOverview

• Fundamentals of CountingFundamentals of Counting• Multiplication PrincipalMultiplication Principal• Permutations: Permutations: ordered arrangementsordered arrangements• Combinations: Combinations: unordered unordered

arrangementsarrangements

• Selected ActivitiesSelected Activities

Counting TechniquesCounting Techniques

• Fundamentals of CountingFundamentals of Counting

• PermutationsPermutations: : ordered arrangementsordered arrangements

• CombinationsCombinations: : unordered arrangementsunordered arrangements

Fundamentals of CountingFundamentals of Counting

• QQ: Jill has 9 shirts and 4 pairs of : Jill has 9 shirts and 4 pairs of pants. How many different pants. How many different outfits does she have?outfits does she have?

• AA: 9 x 4 = 36 : 9 x 4 = 36

36 different outfits36 different outfits


• Multiplication PrincipleMultiplication Principle:: If there are If there are aa ways of choosing one ways of choosing one

thing, and thing, and bb ways of choosing a ways of choosing a second thing after the first is chosen, second thing after the first is chosen, then the total number of choice then the total number of choice patterns is: patterns is:

aa x x bb


• QQ: 3 freshman, 4 sophomores, 5 : 3 freshman, 4 sophomores, 5 juniors, and 2 seniorsjuniors, and 2 seniors are running for are running for SGA representative. One individualSGA representative. One individual will be selected from each class. How will be selected from each class. How many different representative many different representative orderings are possible?orderings are possible?

• AA: 3 · 4 · 5 · 2 = : 3 · 4 · 5 · 2 = 120 different 120 different representative orderingsrepresentative orderings

Generalized Multiplication Generalized Multiplication PrinciplePrinciple

• If there are If there are aa ways of choosing one ways of choosing one thing, thing, bb ways of choosing a second ways of choosing a second thing after the first is chosen, and thing after the first is chosen, and cc ways of choosing a third thing after ways of choosing a third thing after the first two have been chosen…and the first two have been chosen…and zz ways of choosing the last item after ways of choosing the last item after the earlier choices, then the total the earlier choices, then the total number of choice patterns is number of choice patterns is

aa x x bb x x cc x … x x … x zz

ExampleExample

• Q: When I lived in Madison Co., Q: When I lived in Madison Co., AL, the license plates had 2 fixed AL, the license plates had 2 fixed numbers, 2 variable letters and 3 numbers, 2 variable letters and 3 variable numbers. How many variable numbers. How many different license plates were different license plates were possible?possible?

• A: 26 x 26 x 10 x 10 x 10 = A: 26 x 26 x 10 x 10 x 10 = 676,000 different plates676,000 different plates


Q: How many more license plate Q: How many more license plate numbers will Madison County numbers will Madison County gain by changing to 3 letters and gain by changing to 3 letters and 2 numbers?2 numbers?

A: 26 x 26 x 26 x 10 x 10 = A: 26 x 26 x 26 x 10 x 10 = 1,757,600 1,757,600

• 1,757,600 – 676,000 = 1,757,600 – 676,000 = 1,081,600 1,081,600 more license plate numbersmore license plate numbers

Permutations: Permutations: Ordered ArrangementsOrdered Arrangements

• Q: Given 6 people and 6 chairs in a line, Q: Given 6 people and 6 chairs in a line, how many seating arrangements how many seating arrangements (orderings) are possible?(orderings) are possible?

• A: 6 · 5 · 4 · 3 · 2 · 1 = 6! A: 6 · 5 · 4 · 3 · 2 · 1 = 6! = = 720 orderings720 orderings

Permutations: Permutations: Ordered ArrangementsOrdered Arrangements

• Q: Given 6 people and 4 chairs in a line, Q: Given 6 people and 4 chairs in a line, how many different orderings are how many different orderings are possible?possible?

• A: 6 · 5 · 4 · 3 = A: 6 · 5 · 4 · 3 = 360 different orderings360 different orderings

Permutations:Permutations:Ordered ArrangementsOrdered Arrangements

• Permutation of Permutation of nn objects taken objects taken rr at a time: at a time:

rr-permutation, P(-permutation, P(nn,,rr), ), nnPPrr

Q: Given 6 people and 5 chairs in a line, Q: Given 6 people and 5 chairs in a line, how many different orderings are how many different orderings are possible?possible?

A: 6 · 5 · 4 · 3 · 2 = A: 6 · 5 · 4 · 3 · 2 = 720 different orderings720 different orderings


nnPPr r = = nn((nn-1)···(-1)···(nn-(-(rr-1))-1)) = = nn((nn-1)···(-1)···(nn--rr+1)+1) = = nn((nn-1)···(-1)···(nn--rr+1) +1) ((nn--rr)!)! ((nn--rr)!)! = = nn((nn-1)···(-1)···(nn--rr+1)(+1)(nn--rr)···(3)(2)(1))···(3)(2)(1) ((nn--rr)!)! = = nn! ! ((nn--rr)!)!

)!(

!

rn

nPrn


• Q: How many different batting orders are Q: How many different batting orders are possible for a baseball team consisting possible for a baseball team consisting of 9 players?of 9 players?

• A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9!A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9! = = 362,880 batting orders362,880 batting orders

• Note: this is equivalent to Note: this is equivalent to 99PP99.. 99PP99 = = 9! 9! = = 9! 9! = 9! = 9! (9-9)! 0!(9-9)! 0!


• Q: How many different batting Q: How many different batting orders are possible for the leading orders are possible for the leading fourfour batters? batters?

• A: 9 · 8 · 7 · 6 = A: 9 · 8 · 7 · 6 = 3,024 orders 3,024 orders

99PP44 = = 9! 9! = = 9! 9! = = 9! 9!

(9-4)! 5! 5!(9-4)! 5! 5!

Permutations:Permutations:Indistinguishable ObjectsIndistinguishable Objects

• Q: How many different letter Q: How many different letter arrangements can be formed using the arrangements can be formed using the letters letters T E N N E S S E ET E N N E S S E E ? ?

• A: There are 9! Permutations of the A: There are 9! Permutations of the letters letters T E N N E S S E ET E N N E S S E E if the letters if the letters are are distinguishabledistinguishable..

• However, 4 E’s are iHowever, 4 E’s are indistinguishablendistinguishable. . There are 4! ways to order the E’s.There are 4! ways to order the E’s.

Permutations:Permutations:Indistinguishable Objects, Cont.Indistinguishable Objects, Cont.• 2 S’s and 2 N’s are 2 S’s and 2 N’s are indistinguishable. indistinguishable.

There are 2! orderings of each.There are 2! orderings of each.• Once all letters are ordered, there is Once all letters are ordered, there is

only one place for the T.only one place for the T.• If the E’s, N’s, & S’s are If the E’s, N’s, & S’s are

indistinguishableindistinguishable among themselves, among themselves, then there arethen there are

9! 9! = 3,780 different orderings of= 3,780 different orderings of (4!·2!·2!) (4!·2!·2!) T E N N E S S E ET E N N E S S E E

Permutations:Permutations:Indistinguishable ObjectsIndistinguishable Objects

Subsets of Indistinguishable ObjectsSubsets of Indistinguishable Objects

Given Given nn objects of which objects of which aa are alike, are alike, bb are alike, …, and are alike, …, and zz are alike are alike

There are There are nn!! permutations. permutations. a!·b!···z!a!·b!···z!

Combinations:Combinations:Unordered ArrangementsUnordered Arrangements

• Combinations:Combinations: number of different number of different groups of size groups of size rr that can be chosen that can be chosen from a set of from a set of nn objects (order is objects (order is irrelevant)irrelevant)

• Q: From a group of 6 people, select Q: From a group of 6 people, select 4. How many different possibilities 4. How many different possibilities are there?are there?

• A: There are A: There are 66PP44=360 different =360 different orderingsorderings of 4 people out of 6. of 4 people out of 6.

6·5·4·3 = 360 = 6·5·4·3 = 360 = 66PP44 = = n!

(n -r)!

Unordered Example continuedUnordered Example continued

• However the However the orderorder of the chosen 4 people is of the chosen 4 people is irrelevant. There are 24 different orderings irrelevant. There are 24 different orderings of 4 objects.of 4 objects.

4 · 3 · 2 · 1 = 24 = 4! =4 · 3 · 2 · 1 = 24 = 4! =rr!!

• Divide the Divide the total numbertotal number of orderings by the of orderings by the number of number of orderings of the 4 chosen orderings of the 4 chosen people.people. 360360 = = 15 different groups of 4 people15 different groups of 4 people..

2424

Combinations: Combinations: Unordered ArrangementsUnordered Arrangements

The number of ways to choose The number of ways to choose rr objects from a group of objects from a group of n n objectsobjects..

C(n,r)C(n,r) or or nnCCrr, read as “, read as “nn choose choose rr””

)!(!

!

rnr

n

r

nCrn


• Q: From a group of 20 people, a Q: From a group of 20 people, a committee of 3 is to be chosen. committee of 3 is to be chosen. How many different committees How many different committees are possible?are possible?

• A: A:

committeesdifferent 1140)!320(!3

!20

3

20320

C


• Q: From a group of 5 men & 7 women, Q: From a group of 5 men & 7 women, how many different committees of 2 how many different committees of 2 men & 3 women can be found?men & 3 women can be found?

• A: There are A: There are 55CC22 groups of men & groups of men & 77CC33

groups of women. Using the groups of women. Using the multiplication principle multiplication principle

women3 &men 2 of committees possible 350

35035*10)!37(!3

!7*

)!25(!2

!5

3

7*

2

5

Practice ProblemPractice Problem

• You have 30 students in your class, You have 30 students in your class, which will be arranged in 5 rows of which will be arranged in 5 rows of 6 people. Assume that any student 6 people. Assume that any student can sit in any seat. can sit in any seat. • How many different seating charts How many different seating charts

could you have for the first row?could you have for the first row?• How many different seating charts How many different seating charts

could you have for the whole class?could you have for the whole class?

It’s Your TurnIt’s Your Turn

• Make up three counting problems Make up three counting problems which would interest your students, which would interest your students, include one permutation and one include one permutation and one combination and one of your combination and one of your choice. choice.

• Calculate the answer for these Calculate the answer for these problems.problems.

Questions ???Questions ???

• Fundamentals of Counting?Fundamentals of Counting?

• Permutations?Permutations?

• Combinations?Combinations?

HomeworkHomework

Documents

Counting Fundamentals