sets and set operations the number of elements in a finite set the multiplication principle

66

Sets and Set OperationsSets and Set Operations The Number of Elements in a Finite SetThe Number of Elements in a Finite Set The Multiplication Principle The Multiplication Principle Permutations and CombinationsPermutations and Combinations

Sets and CountingSets and Counting

6.16.1Sets and Set OperationsSets and Set Operations

A

C

B

U

Set Terminology and NotationSet Terminology and Notation

A A setset is a well-defined is a well-defined collection of objectscollection of objects..✦ Sets are usually denoted by Sets are usually denoted by upper caseupper case

letters such as letters such as AA, , BB, , CC, …, … The The objectsobjects of a set are called of a set are called elementselements, or , or

membersmembers, of a set., of a set.✦ Elements are usually denoted by Elements are usually denoted by lower caselower case

letters such as letters such as aa, , bb, , cc, …, …


The elements of a set may be displayed by The elements of a set may be displayed by listing listing each element between braceseach element between braces..

For example, using For example, using roster notationroster notation, the set , the set AA consisting of the first three letters of the English consisting of the first three letters of the English alphabet is writtenalphabet is written

AA = { = {aa, , bb, , cc}}

The set The set BB of all letters of the alphabet may be of all letters of the alphabet may be writtenwritten

BB = { = {aa, , bb, , cc, …,, …, z z}}


Another notation commonly used is Another notation commonly used is set-builder set-builder notationnotation..

Here, a Here, a rulerule is given that is given that describesdescribes the the definite definite propertyproperty or properties an or properties an objectobject xx must satisfy to must satisfy to qualify for membership in the set.qualify for membership in the set.

For example, the set For example, the set BB of all letters of the alphabet of all letters of the alphabet may be written asmay be written as

BB = { = {x x | | xx is a letter of the English alphabetis a letter of the English alphabet}}

and is and is readread “ “BB is the set of all elements of is the set of all elements of x x such such that that xx is a letter of the English alphabet.” is a letter of the English alphabet.”


There is also terminology regarding to whether an There is also terminology regarding to whether an elementelement belongsbelongs to a set or not: to a set or not:✦ If If aa is an is an element ofelement of a set a set AA, we write , we write aa AA and and

read “read “aa belongs tobelongs to AA” or “” or “aa is an is an element ofelement of AA.”.”

✦ If If aa is is notnot an an element ofelement of a set a set AA, we write , we write aa AA and read “and read “aa does does notnot belong to belong to AA” or “” or “aa is is notnot an an element ofelement of AA.”.”

Set EqualitySet Equality

Two sets Two sets A A andand B B are are equalequal, written , written A A == B B, , if and only if they have exactly the if and only if they have exactly the same same elementselements..

ExampleExample

Let Let AA, , BB, and , and CC be the sets be the sets

Then, Then, A A = = BB since they both contain exactly the since they both contain exactly the same elementssame elements..

Note that the Note that the orderorder in which the elements are in which the elements are displayed is displayed is immaterialimmaterial..

Also, Also, AA ≠ ≠ CC since since uu AA but but uu CC.. Similarly, we conclude that Similarly, we conclude that BB ≠ ≠ CC..

{ , , , , }

{ , , , , }

{ , , , }

A a e i o u

B a i o e u

C a e i o

{ , , , , }

{ , , , , }

{ , , , }

A a e i o u

B a i o e u

C a e i o

Example 1, page 314Example 1, page 314

SubsetSubset

If every element of a set If every element of a set AA is also an element is also an element of a set of a set BB, then we say that , then we say that AA is a is a subsetsubset of of BB and write and write AA BB..

By this definition, two sets By this definition, two sets AA and and BB are are equalequal if and only if if and only if (1)(1) A A BB and and (2)(2) BB AA..

ExampleExample

Consider again the sets Consider again the sets AA, , BB, and , and CC

We find that We find that C C BB since every element of since every element of CC is is also an element of also an element of BB..

Also, Also, AA is not a subset of is not a subset of CC, written , written A A CC, , since since uu AA but but uu CC. .

{ , , , , }

{ , , , , }

{ , , , }

A a e i o u

B a i o e u

C a e i o

{ , , , , }

{ , , , , }

{ , , , }

A a e i o u

B a i o e u

C a e i o


Empty SetEmpty Set

The set that The set that containscontains no elementsno elements is called is called the the empty setempty set and is denoted by and is denoted by ØØ..

The empty set, The empty set, ØØ, is a , is a subset of every setsubset of every set. . To see this, observe that To see this, observe that ØØ has no elements has no elements

and thus contains no element that is not and thus contains no element that is not also in also in AA..

ExampleExample

List all subsets of the set List all subsets of the set AA = { = {aa, , bb, , cc}}..

SolutionSolution There is There is one subsetone subset containing containing no elementsno elements, , ØØ.. There are There are three subsetsthree subsets consisting of consisting of one elementone element: :

{{aa}, {}, {bb}, {}, {cc}} There are also There are also three subsetsthree subsets consisting of consisting of two elementstwo elements::

{{aa,, b b}, {}, {aa,, c c}, {}, {bb, , cc}} Finally there is Finally there is one subsetone subset consisting of consisting of three elementsthree elements, ,

the set the set AA itself. itself. Therefore, the subsets of Therefore, the subsets of AA are are

ØØ, {, {aa}, {}, {bb}, {}, {cc}, {}, {aa,, b b}, {}, {aa,, c c}, {}, {bb, , cc}, {}, {aa, , bb, , cc}}


Universal SetUniversal Set

A A universal setuniversal set is the is the set of all elements of set of all elements of interestinterest in a particular discussion. in a particular discussion.

It is the It is the largest setlargest set in the sense that in the sense that all setsall sets considered in the discussion of the problem considered in the discussion of the problem are subsetsare subsets of the universal set. of the universal set.

ExampleExample

If the problem at hand is to determine the If the problem at hand is to determine the ratio of female ratio of female to male students in a collegeto male students in a college, then the logical choice of a , then the logical choice of a universal setuniversal set is the set consisting of the is the set consisting of the whole student body whole student body of the collegeof the college..

If the problem is to determine the If the problem is to determine the ratio of female to male ratio of female to male students in the business departmentstudents in the business department of the college, then the of the college, then the set of set of all students in the business departmentall students in the business department may be may be chosen as the chosen as the universal setuniversal set..


Set UnionSet Union

Let Let AA and and BB be sets. The be sets. The unionunion of of AA and and BB, , written written A A BB, is the set of all elements that , is the set of all elements that belong to either belong to either AA or or BB or both. or both.

A A BB = { = {x x | | x x A A oror x x B B or bothor both}}

ExampleExample

If If AA = { = {bb, , c, d, ec, d, e}} and and BB = { = {aa, , bb, , cc, , dd}}, ,

then then A A BB = { = {aa, , bb, , cc, , d, ed, e}}..


AA BB

eebbccdd

aa

Set IntersectionSet Intersection

Let Let AA and and BB be sets. The set of be sets. The set of elements common elements common toto with the sets with the sets AA and and BB, written , written A A BB, is called , is called the the intersectionintersection of of AA and and BB..

A A BB = { = {x x | | x x A A andand x x BB}}

ExampleExample

If If AA = { = {bb, , c, d, ec, d, e}} and and BB = { = {aa, , bb, , cc, , dd}}, ,

then then A A BB = { = {bb, , c, dc, d}}..


AA BB

bbccdd

Complement of a SetComplement of a Set

If If UU is a is a universal setuniversal set and and AA is a is a subsetsubset of of UU, , then the set of all elements in then the set of all elements in UU that are that are notnot in in A A is called the is called the complementcomplement of of AA and is and is denoted denoted AAcc..

AAcc = {= {x x | | x x UU and and x x AA}}

ExampleExample

If If UU = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

and and AA = {2, 4, 6, 8, 10} = {2, 4, 6, 8, 10}, ,

then then AAcc = {1, 3, 5, 7, 9}= {1, 3, 5, 7, 9}..

AA

UU

AAcc

Set ComplementationSet Complementation

If If UU is a universal set and is a universal set and AA is a subset of is a subset of UU, , then then

a.a. UUcc = = ØØ b.b.ØØcc = = UU

c.c. ((AAcc))cc = = A A d.d. A A AAcc = = UU

e.e.A A AAcc = = ØØ

Properties of Set OperationsProperties of Set Operations

Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Commutative law for unionCommutative law for union

A A BB = = B B AA


Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Commutative law for intersectionCommutative law for intersection

A A BB = = B B AA


Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Associative law for unionAssociative law for union

A A ((B B C C) = () = (A A BB)) CC


Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Associative law for intersectionAssociative law for intersection

A A ((B B C C) = () = (A A BB)) CC


Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Distributive law for unionDistributive law for union

A A ((B B C C) = () = (A A BB)) ((A A CC))


Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ Distributive law for intersectionDistributive law for intersection

A A ((B B C C) = () = (A A BB)) ((A A CC))

Set OperationsSet Operations

Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ De Morgan’s LawsDe Morgan’s Laws

((A A B B))cc = = AAcc BBcc

Set OperationsSet Operations

Let Let UU is a universal set. If is a universal set. If AA , , BB, and , and CC are are arbitrary subsets of arbitrary subsets of UU, then we have the , then we have the following laws:following laws:✦ De Morgan’s LawsDe Morgan’s Laws

((A A B B))cc = = AAcc B Bcc

ExampleExample

LetLet UU = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

AA = {1, 2, 4, 8, 9}= {1, 2, 4, 8, 9}

BB = {3, 4, 5, 6, 8}= {3, 4, 5, 6, 8}

Verify by direct computation that Verify by direct computation that ((A A B B))cc = = AAcc BBcc..SolutionSolution

A A B B = {1, 2, 3, 4, 5, 6, 8, 9} = {1, 2, 3, 4, 5, 6, 8, 9}, so , so ((A A B B))cc = {7, 10} = {7, 10}..

Moreover, Moreover, AAcc = {3, 5, 6, 7, 10}= {3, 5, 6, 7, 10} and and BBcc = {1, 2, 7, 9, 10}= {1, 2, 7, 9, 10}, ,

so so AAcc BBcc = {7, 10}= {7, 10}..


Applied Example:Applied Example: Automobile Options Automobile Options Let Let UU denote the set of denote the set of all cars in a dealer’s lotall cars in a dealer’s lot, and let, and let

AA = {= {xx UU || xx is equipped with automatic transmissionis equipped with automatic transmission}}BB = {= {xx UU || xx is equipped with air conditioningis equipped with air conditioning}}

CC = {= {xx UU || xx is equipped with side air bagsis equipped with side air bags}}

Find an expression in terms of Find an expression in terms of AA, , BB, and , and CC for each of the for each of the following sets:following sets:✦ The set of cars with The set of cars with at least oneat least one of the given options. of the given options.

✦ The set of cars with The set of cars with exactly oneexactly one of the given options. of the given options.

✦ The set of cars withThe set of cars with automatic transmission automatic transmission andand side air side air bags bags butbut nono air conditioning air conditioning..

Applied Example 14, page 319Applied Example 14, page 319




SolutionSolution The set of cars with The set of cars with at least oneat least one of the given options is of the given options is

given by given by A A B B C C::

AABB

CC

UU





SolutionSolution The set of cars with The set of cars with exactly oneexactly one of the given options is of the given options is

given by given by ((A A B B cc CC

cc ) ) ((B B C C

cc AA cc

) ) ((C C A A cc BB

cc ))::

AA

CC

BB

UU





SolutionSolution The set of cars withThe set of cars with automatic transmission automatic transmission andand side air side air

bags bags butbut nono air conditioning air conditioning is given by is given by A A C C BB cc::

AABB

CC

UU


6.26.2The Number of Elements in a Finite SetThe Number of Elements in a Finite Set

nn((ØØ)) = 0= 0

nn((A A B B)) = = nn((AA) + ) + nn((BB) – ) – nn((AA BB))

nn((A A B B C C)) = = nn((AA) + ) + nn((BB) + ) + nn((CC))

– – nn((AA BB) – ) – nn((AA CC) – ) – nn((BB CC) )

+ + nn((AA B B CC) )

Counting Elements of a SetCounting Elements of a Set

The The number of elementsnumber of elements in a finite set is determined by in a finite set is determined by simply simply counting the elementscounting the elements in the set. in the set.

If If AA is a set, then is a set, then nn((AA) ) denotes the denotes the number of elementsnumber of elements in in AA. .

For example, ifFor example, if

AA = {1, 2, 3, …,20} = {1, 2, 3, …,20} BB = {a, b} = {a, b} CC = {8} = {8}

then then nn((AA) = 20) = 20, , nn((BB) = 2) = 2, and , and nn((CC) = 1) = 1..

The The empty setempty set has has no elementsno elements in it, so in it, so nn((ØØ) = 0) = 0..

If If AA and and BB are are disjoint setsdisjoint sets, then, then

nn((A A B B) = ) = nn((AA) + ) + nn((BB))

ExampleExample If If AA = { = {aa, , cc, , dd} } andand BB = { = {bb, , ee, , ff, , gg}}, then, then

we see that we see that A A B B = = {{aa, , bb, , cc, , dd,, e e, , ff, , gg} } so,so, nn((A A B B) = 7) = 7..

On the other hand, On the other hand, nn((AA) + ) + nn((BB) = 3 + 4 = 7) = 3 + 4 = 7..

Thus, Thus, nn((A A B B) = ) = nn((AA) + ) + nn((BB) ) holds true holds true in this casein this case, , because because AA and and BB are are disjoint setsdisjoint sets: : AA BB = = ØØ..



In the general case, In the general case, AA and and BB need not beneed not be disjointdisjoint, which , which leads us to the formulaleads us to the formula

nn((A A B B) = ) = nn((AA) + ) + nn((BB) – ) – nn((AA BB) )

ExampleExample

If If AA = { = {aa, , bb, , cc, , d d , , ee} } and and BB = { = {bb, , dd, , ff, , hh}}, ,

then, then, A A B B = = {{aa, , bb, , cc, , dd,, e e, , ff, , hh} } so so nn((A A B B) = 7) = 7..

On the other hand, On the other hand, ((AA BB) = {) = {bb, , dd}}, so , so nn((A A B B) = 2) = 2.. We also see that We also see that nn((AA) = 5 ) = 5 and and nn((BB) = 4) = 4.. Thus, we find that Thus, we find that the formula holds truethe formula holds true::

nn((AA) + ) + nn((BB) – ) – nn((AA BB) = 5 + 4 – 2 = 7 = ) = 5 + 4 – 2 = 7 = nn((A A B B))



Applied Example: Applied Example: Consumer SurveysConsumer Surveys

In a survey of In a survey of 100100 coffee drinkers, it was found that coffee drinkers, it was found that 7070 take take sugarsugar, , 6060 take take creamcream, and , and 5050 take take bothboth sugarsugar and and creamcream with with their coffee. How many coffee drinkers take their coffee. How many coffee drinkers take sugarsugar,, cream cream, or, or bothboth with their coffee? with their coffee?

SolutionSolution Let Let UU be the set of be the set of 100100 coffee drinkers surveyed, and let coffee drinkers surveyed, and let

A A = {= {xx ∈ ∈ UU || xx takes sugar} takes sugar} andand B B = {= {xx ∈ ∈ UU || xx takes cream} takes cream}.. Then, Then, nn((AA) = 70) = 70,, nn((BB) = 60) = 60, and , and nn((AA BB) = 50) = 50.. The set of coffee drinkers who take The set of coffee drinkers who take sugarsugar,, cream cream, or, or both both

with their coffee is given by with their coffee is given by A A B B.. Using the Using the formulaformula for counting elements of a for counting elements of a union of not union of not

disjoint setsdisjoint sets, we find, we find

nn((A A B B) = ) = nn((AA) + ) + nn((BB) – ) – nn((AA BB) = 70 +60 – 50 = 80) = 70 +60 – 50 = 80

Thus, Thus, 8080 out of the out of the 100100 surveyed coffe drinkers take surveyed coffe drinkers take sugarsugar,, creamcream, or, or both both with their coffee. with their coffee.


Similar rules can be derived for cases involving Similar rules can be derived for cases involving more than more than two setstwo sets. .

For example, if we have For example, if we have three setsthree sets AA, , BB, and , and CC, we find that, we find that

nn((A A B B C C) = ) = nn((AA) + ) + nn((BB) + ) + nn((CC) ) – – nn((AA BB) )

– – nn((AA CC) – ) – nn((BB CC) + ) + nn((AA B B CC) )


6.36.3The Multiplication PrincipleThe Multiplication Principle

a (I, a)

b (I, b)

a (II, a)

b (II, b)

a (III, a)

b (III, b)

I

II

III

A

BC

The Fundamental Principle of CountingThe Fundamental Principle of Counting

The Multiplication PrincipleThe Multiplication Principle Suppose there are Suppose there are mm ways of performing a ways of performing a

tasktask TT11 and and n n waysways of performing a of performing a tasktask TT22..

Then, there are Then, there are mnmn waysways of performing the of performing the tasktask TT11 followedfollowed by the by the tasktask TT22..

Three Three trunk roads connect trunk roads connect town Atown A and and town Btown B, and , and twotwo trunk roads connect trunk roads connect town Btown B and and town Ctown C..

Use the multiplication principle to find the Use the multiplication principle to find the number of number of waysways a journey from a journey from town Atown A to to town Ctown C via via town Btown B may be may be completed.completed.

SolutionSolution There are There are three waysthree ways of performing the of performing the first taskfirst task: :

✦ Going from Going from town Atown A to to town Btown B.. Then, there are Then, there are two waystwo ways of performing the of performing the second tasksecond task: :

✦ Going from Going from town Btown B to to town Ctown C.. The The multiplication principlemultiplication principle says there are says there are 3 3 ·· 2 = 6 2 = 6 ways ways

to complete a journey from to complete a journey from town Atown A to to town Ctown C via via town Btown B..

ExampleExample


ExampleExample



SolutionSolution The The six wayssix ways to perform the desired task can be seen in the to perform the desired task can be seen in the

diagram below:diagram below:

The The six possible routessix possible routes are: are:(I, (I, aa),), (I, (I, bb),), (II, (II, aa),), (II, (II, bb),), (III, (III, aa),), (III, (III, bb))

AA

II

IIII

IIIIII

CC

aa

bb

BB


ExampleExample



SolutionSolution They can also be seenThey can also be seen

with the aid of a with the aid of a tree diagramtree diagram::

aa (I, (I, aa))

bb (I, (I, bb))

aa (II, (II, aa))

bb (II, (II, bb))

aa (III, (III, aa))

bb (III, (III, bb))

II

IIII

IIIIII

AA

BBCC


Generalized Multiplication PrincipleGeneralized Multiplication Principle

Suppose a Suppose a tasktask TT11 can be performed in can be performed in NN11 waysways, ,

a a tasktask TT22 can be performed in can be performed in NN22 waysways, …, and, , …, and,

finally, a finally, a tasktask TTnn can be performed in can be performed in NNnn waysways..

Then, the Then, the number number ofof waysways of performing of performing taskstasks TT11, , TT22, …, , …, TTnn in successionin succession is given by the is given by the

productproduct

NN1 1 NN22 ······ NNnn

ExampleExample

A coin is tossed A coin is tossed 33 times, and the sequence of times, and the sequence of headsheads and and tailstails is recorded. is recorded.

Use the Use the generalized multiplication principlegeneralized multiplication principle to determine to determine the the numbernumber of of possible outcomespossible outcomes of this activity. of this activity.

SolutionSolution The coin may land in The coin may land in two waystwo ways.. Therefore, in Therefore, in three tossesthree tosses the number of the number of outcomesoutcomes

(sequences) is given by (sequences) is given by

2 2 · · 2 2 · · 2 2 = 8= 8


Applied Example:Applied Example: Combination Locks Combination Locks

A combination lock is unlocked by dialing a sequence of A combination lock is unlocked by dialing a sequence of numbers:numbers:✦ First to the First to the leftleft, then to the , then to the rightright, and to the , and to the leftleft again. again.

If there are If there are ten digitsten digits on the dial, determine the on the dial, determine the number of number of possible combinationspossible combinations..

SolutionSolution There are There are ten choicesten choices for the for the first numberfirst number, followed by , followed by tenten

for the for the secondsecond and and tenten for the for the thirdthird, so by the generalized , so by the generalized multiplication principle there are multiplication principle there are

10 10 · · 10 10 · · 10 10 = 1000= 1000

possible combinations.possible combinations.


6.46.4Permutations and CombinationsPermutations and Combinations

1!1! = 1= 1

2!2! = 2 = 2 ·· 1 = 21 = 2

3!3! = 3 = 3 ·· 2 2 ·· 1 = 61 = 6

4!4! = 4 = 4 ·· 3 3 ·· 2 2 ·· 1 = 241 = 24

5!5! = 5 = 5 ·· 4 4 ·· 3 3 ·· 2 2 ·· 1 = 1201 = 120 . . .. ..

10!10! = 10 = 10 ·· 9 9 ·· 8 8 ·· 7 7 ·· 6 6 · · 5 5 ·· 4 4 ·· 3 3 ·· 2 2 ·· 1 = 3,628,8001 = 3,628,800

PermutationsPermutations

Given Given a set of distinct objectsa set of distinct objects, a , a permutationpermutation of the set is of the set is an an arrangementarrangement of these objects in a of these objects in a definite orderdefinite order..

Order matters very often in the real world.Order matters very often in the real world. For example, suppose the For example, suppose the winning numberwinning number for the first for the first

prize in a prize in a raffleraffle is is 92379237.. Then, the number Then, the number 29732973 cannot be the winnercannot be the winner, even though , even though

it contains the it contains the same digitssame digits as as 92379237.. Here, the Here, the set of the four numbersset of the four numbers 99, , 22, , 33, and , and 77 are are

arrangedarranged in a different order: one arrangement results in in a different order: one arrangement results in a a prize winnerprize winner, the other does , the other does notnot..

ExampleExample

Find the number of permutations of Find the number of permutations of A A = {= {aa,, b b,, c c}}..SolutionSolution Each Each permutationpermutation of of AA consists of a consists of a sequence sequence of the three of the three

letters letters aa, , bb, , cc.. Any such Any such sequencesequence can be constructed by filling in each of can be constructed by filling in each of

the the three blanksthree blanks below with below with oneone of the of the three lettersthree letters::

Now, there are Now, there are three lettersthree letters with which we may fill the with which we may fill the first blankfirst blank: : aa, , bb, or , or cc..

Having selected a letter for the first blank, there are Having selected a letter for the first blank, there are two two lettersletters left for the left for the second blanksecond blank..

Finally, there is but Finally, there is but one letterone letter left to fill the left to fill the third blankthird blank.. Invoking the Invoking the generalized multiplication principlegeneralized multiplication principle, we see , we see

that there are that there are 3 3 · · 2 2 · · 1 = 61 = 6 possible permutations of possible permutations of AA..Example 1, page 336Example 1, page 336

ExampleExample

Find the number of permutations of Find the number of permutations of A A = {= {aa,, b b,, c c}}..SolutionSolution The The tree diagramtree diagram shows the shows the 66 possible possible permutationspermutations of of AA::

bb ((aa, , bb, , cc))

cc ((aa, , cc, , bb))

aa ((bb, , aa, , cc))

cc ((bb, , cc, , aa))

aa ((cc, , aa, , bb))

bb ((cc, , bb, , aa))

aa

bb

cc

Combined Combined outcomesoutcomes


nn-Factorial-Factorial

For any natural number For any natural number nn,,

nn!! = = nn((nn – 1)( – 1)(nn – 2) – 2) ·· · · · · ·· 3 3 ·· 2 2 ·· 1 1

0!0! = 1= 1

ExampleExample

1!1! = 1= 1

2!2! = 2 = 2 ·· 1 = 21 = 2

3!3! = 3 = 3 ·· 2 2 ·· 1 = 61 = 6

4!4! = 4 = 4 ·· 3 3 ·· 2 2 ·· 1 = 241 = 24

5!5! = 5 = 5 ·· 4 4 ·· 3 3 ·· 2 2 ·· 1 = 1201 = 120 . . .. ..

10!10! = 10 = 10 ·· 9 9 ·· 8 8 ·· 7 7 ·· 6 6 · · 5 5 ·· 4 4 ·· 3 3 ·· 2 2 ·· 1 = 3,628,8001 = 3,628,800

Permutations ofPermutations of n n Distinct ObjectsDistinct Objects

The number of permutations of The number of permutations of nn distinct distinct objects taken objects taken rr at a time is at a time is

!( , )

( )!

nP n r

n r

!

( , )( )!

nP n r

n r

ExampleExample

Compute Compute PP(4, 4)(4, 4) and and PP(4, 2)(4, 2) and interpret the results. and interpret the results.

SolutionSolution

The number of The number of permutationspermutations of of four objectsfour objects taken taken four at four at a timea time is is 2424..

!( , )

( )!

nP n r

n r

!

( , )( )!

nP n r

n r

4! 4! 4 3 2 1(4,4) 24

(4 4)! 0! 1P

4! 4! 4 3 2 1

(4,4) 24(4 4)! 0! 1

P


ExampleExample

Compute Compute PP(4, 4)(4, 4) and and PP(4, 2)(4, 2) and interpret the result. and interpret the result.

SolutionSolution

The number of The number of permutationspermutations of of four objectsfour objects taken taken two at a two at a timetime is is 1212..

!( , )

( )!

nP n r

n r

!

( , )( )!

nP n r

n r

4! 4! 4 3 2 1(4,2) 12

(4 2)! 2! 2 1P

4! 4! 4 3 2 1

(4,2) 12(4 2)! 2! 2 1

P


Applied Example:Applied Example: Committee Selection Committee Selection

Find the Find the number of waysnumber of ways a chairman, a vice-chairman, a a chairman, a vice-chairman, a secretary, and a treasurer can be chosen from a secretary, and a treasurer can be chosen from a committeecommittee of of eight memberseight members..

SolutionSolution The problem is equivalent to finding the The problem is equivalent to finding the numbernumber of of

permutationspermutations of of eighteight distinct objects taken distinct objects taken fourfour at a time. at a time. Therefore, there areTherefore, there are

ways of choosing the ways of choosing the fourfour officials from the committee of officials from the committee of eighteight members. members.

8! 8!(8,4) 8 7 6 5 1680

(8 4)! 4!P

8! 8!

(8,4) 8 7 6 5 1680(8 4)! 4!

P


Permutations ofPermutations of n n Objects, Not All DistinctObjects, Not All Distinct

Given a set of Given a set of nn objects in which objects in which nn11 objects are objects are

alike and of one kind, alike and of one kind, nn22 objects are alike and of objects are alike and of

another kind, …, and another kind, …, and nnrr objects are alike and of objects are alike and of

yet another kind, so thatyet another kind, so that

then the number of permutations of these then the number of permutations of these nn objects taken objects taken nn at a time is given by at a time is given by

1 2

!

! ! !r

n

n n n1 2

!

! ! !r

n

n n n

1 2 rn n n n 1 2 rn n n n

ExampleExample

Find the number of Find the number of permutationspermutations that can be formed from that can be formed from all the letters in the word all the letters in the word ATLANTAATLANTA..

SolutionSolution There are There are seven objectsseven objects (letters) involved, so (letters) involved, so nn = 7 = 7.. ThreeThree of them are of them are alikealike and of and of one kindone kind ( (threethree AAs), while s), while

twotwo of them are of them are alikealike and of and of another kindanother kind (the (the twotwo TTs).s). Hence, in this case we have Hence, in this case we have nn11 = 3 = 3, , nn22 = 2 = 2, , nn33 = 1 = 1, and , and nn44 = 1 = 1..

Therefore, there areTherefore, there are

possiblepossible permutations permutations..

1 2

! 7! 7 6 5 4 3 2 1420

! ! ! 3!2!1!1! 3 2 1 2 1 1 1r

n

n n n

1 2

! 7! 7 6 5 4 3 2 1420

! ! ! 3!2!1!1! 3 2 1 2 1 1 1r

n

n n n


Applied Example:Applied Example: Management Decision Management Decision

Weaver and Kline, a stock brokerage firm, has received Weaver and Kline, a stock brokerage firm, has received ninenine inquiriesinquiries regarding new accounts. regarding new accounts.

In how many ways can these inquiries be directed to In how many ways can these inquiries be directed to threethree of the firm’s of the firm’s accountaccount executivesexecutives if each account executive if each account executive is to handle is to handle threethree inquiriesinquiries??

SolutionSolution If we think of the If we think of the 99 inquiries as being slots arranged in a inquiries as being slots arranged in a

row with row with inquiryinquiry 1 1 on the left and on the left and inquiryinquiry 99 on the right, on the right, then the problem can be thought of as one of filling each then the problem can be thought of as one of filling each slot with a business card from an account executive.slot with a business card from an account executive.

We will use We will use 99 business cards, of which business cards, of which 33 are alike and of are alike and of one kindone kind, , 33 are alike and of are alike and of another kindanother kind, and , and 33 are alike are alike and of and of yet another kindyet another kind..


Applied Example:Applied Example: Management Decision Management Decision

Weaver and Kline, a stock brokerage firm, has received Weaver and Kline, a stock brokerage firm, has received ninenine inquiriesinquiries regarding new accounts. regarding new accounts.

In how many ways can these inquiries be directed to In how many ways can these inquiries be directed to threethree of the firm’s of the firm’s accountaccount executivesexecutives if each account executive if each account executive is to handle is to handle threethree inquiriesinquiries??

SolutionSolution In this case we have In this case we have nn = 9 = 9 and and nn11 = = nn22 = = nn33 = 3 = 3..

Therefore, there areTherefore, there are

possiblepossible ways ways of directing the of directing the inquiriesinquiries..

1 2

! 9! 9 8 7 6 5 4 3 2 11680

! ! ! 3!3!3! 3 2 1 3 2 1 3 2 1r

n

n n n

1 2

! 9! 9 8 7 6 5 4 3 2 11680

! ! ! 3!3!3! 3 2 1 3 2 1 3 2 1r

n

n n n


Combinations of Combinations of nn Objects Objects

The number of combinations of The number of combinations of nn distinct distinct objects taken objects taken r r at a time is given byat a time is given by

!( , )

!( )!

nC n r

r n r

!

( , )!( )!

nC n r

r n r

(where(where r r nn))

ExampleExample

Compute Compute CC(4, 4)(4, 4) and and CC(4, 2) (4, 2) and interpret the results.and interpret the results.

SolutionSolution

This gives This gives 11 as the number of as the number of combinationscombinations of of fourfour distinct distinct objectsobjects taken taken fourfour at a time. at a time.

!( , )

!( )!

nC n r

r n r

!

( , )!( )!

nC n r

r n r

4! 4!(4,4) 1

4!(4 4)! 4!0!C

4! 4!

(4,4) 14!(4 4)! 4!0!

C


ExampleExample

Compute Compute CC(4, 4)(4, 4) and and CC(4, 2) (4, 2) and interpret the results.and interpret the results.

SolutionSolution

This givesThis gives 6 6 as the number of as the number of combinationscombinations of of fourfour distinct distinct objectsobjects taken taken twotwo at a time. at a time.

!( , )

!( )!

nC n r

r n r

!

( , )!( )!

nC n r

r n r

4! 4! 4 3(4,2) 6

2!(4 2)! 2!2! 2C

4! 4! 4 3

(4,2) 62!(4 2)! 2!2! 2

C


Applied Example:Applied Example: Committee Selection Committee Selection

A Senate investigation subcommittee of A Senate investigation subcommittee of fourfour members is to members is to be selected from a Senate committee of be selected from a Senate committee of tenten members. members.

Determine the Determine the number of waysnumber of ways this can be done. this can be done.

SolutionSolution In this case, the In this case, the orderorder in which the members are selected is in which the members are selected is

unimportantunimportant and so the number of ways of choosing the and so the number of ways of choosing the subcommittee is given by subcommittee is given by CC(10, 4)(10, 4). .

Hence, there areHence, there are

ways of choosing the subcommittee.ways of choosing the subcommittee.

10! 10! 10 9 8 7(10,4) 210

4!(10 4)! 4!6! 4 3 2 1C

10! 10! 10 9 8 7

(10,4) 2104!(10 4)! 4!6! 4 3 2 1

C


Suppose an investor has decided to purchase shares in Suppose an investor has decided to purchase shares in the stocks of the stocks of twotwo aerospaceaerospace companies, companies, twotwo energyenergy development companies, and development companies, and twotwo electronicselectronics companies. companies.

In In how many wayshow many ways can the investor can the investor selectselect the group of the group of sixsix companies for the investment from a companies for the investment from a recommended recommended listlist of of fivefive aerospaceaerospace companies, companies, threethree energyenergy development companies, and development companies, and fourfour electronicselectronics companies?companies?

Applied Example:Applied Example: Investment Options Investment Options


Applied Example:Applied Example: Investment Options Investment Options

SolutionSolution There are There are CC(5, 2)(5, 2) ways to select the ways to select the aerospaceaerospace companies, companies,

CC(3, 2)(3, 2) ways to select the ways to select the energyenergy development companies, development companies, and and CC(4, 2)(4, 2) ways to select the ways to select the electronicselectronics companies as companies as investments.investments.

By the By the generalized multiplication principlegeneralized multiplication principle, there are, there are

ways of selecting the group of ways of selecting the group of sixsix companiescompanies..

5! 3! 4!(5,2) (3,2) (4,2)

2!3! 2!1! 2!2!5 4 4 3

3 1802 2

C C C

5! 3! 4!(5,2) (3,2) (4,2)

2!3! 2!1! 2!2!5 4 4 3

3 1802 2

C C C


End of End of Chapter Chapter

sets and set operations the number of elements in a finite set the multiplication principle

Documents