unit 2 - lesson 6

MDM4U-B

Organized Counting and Permutations

6

Mathematics of Data Management MDM4U-B Lesson 6 1

IntroductionIn this lesson, you will learn about organized counting. Organized counting is a type of combinatorics. Combinatorics is the branch of mathematics that deals with counting the number of ways in which something can be done in order to show you what the possible outcomes are. Combinatorics has many applications that include scheduling (flights, deliveries, timetables, and so on), finding the most efficient way to route Internet traffic, and working out the probabilities in games of chance, such as lotteries.

Combinatorics is also an important foundation for many other statistical and probability analyses that you will learn about in later units.

Here are some examples of problems that you will be able to solve in this course, using organized counting techniques:

Ifatrackteamconsistsof12members,howmanywayscanthecoachchoosefourofthe members to fill the four different positions in a relay race?

Ifanicecreamstoresells31differentflavoursoficecream,howmanywayscanyouorder a cone that contains three scoops of ice cream?

Insomelotterygames,youchoosesixnumbersfromthenumbers1through49.Howmany different possible lottery tickets are there?

Planning Your StudyYou may find this time grid helpful in planning when and how you will work through this lesson.

Suggested Timing for This Lesson (Hours)Multiplicative Counting Principle Permutations Ordered N-tuples 1Factorial Notation and Permutations 1Key Questions 1

What You Will Learn After completing this lesson, you will be able to

usetheappropriatecountingtechniquestofindthenumberofoutcomesofagivenexperiment or situation

usefactorialstosimplifytheprocessofsolvingcountingproblems

Copyright 2009 The Ontario Educational Communications Authority. All rights reserved.www.ilc.org

Lesson 6 Mathematics of Data Management MDM4U-B2

Multiplicative Counting Principle Combinatorics is the branch of mathematics that deals with calculating the number of possible outcomes in situations. This tells you what is possible. Once you know what is possible, statistics can be used to determine the likelihood that a particular outcome will occur out of all of the possible outcomes. As you will see later in this course, many statistical analyses (for example, binomial probability) are based on these concepts. Thats why it is important to learn something about them now.

One of the main rules you will use in solving counting problems is the multiplicative counting principle. First, take a look at an example to illustrate this important principle.

Suppose that Nora wants to do her grocery shopping. She likes to shop at the farmers market for fresh local produce, and the grocery store for household staples such as rice, diapers, and detergent. She also likes to shop at the farmers market first, because she doesnt want to have to carry heavy grocery items like laundry detergent around the market. Suppose that there are three different routes from her home to the market, three routes from the market to the grocery store, and two routes from the grocery store to her home. How many routes can Nora take to do her shopping and return home?

You can use a diagram to help you count the possible number of routes. This diagram is called a tree diagram. Each line represents a route or branch. For example, after Noras market stop, there are three different routes to the grocery store. One of the branches through Noras tree diagram shown here has been highlighted, to demonstrate one of her possible shopping routes.

House Market

Market

Market

Grocery store

Grocery store

Grocery store

Grocery store

Grocery store

Grocery store

Grocery store

Grocery store

Grocery store

House

House

House

House

House

House

House

House

House

House

HouseHouse

House

House

House

House

House

House

Figure 6.1: Tree diagram showing Noras potential shopping routes

Copyright 2009 The Ontario Educational Communications Authority. All rights reserved. www.ilc.org


The number of routes corresponds to the number of times the word House appears on theright-handsideofthediagram.Ifyoucountthem,yougetatotalof18routes.

Write down the number of routes for each portion or stage of Noras shopping trip:

Housetomarket=3

Markettogrocerystore=3

Grocerystoretohouse=2

ThetotalnumberofroutesthatNoracantakeis332=18.

Suppose that the number of options is too big to draw out in a tree diagram and instead, you want to use a general rule to count the number of ways of performing a task. Suppose also that in order to perform the task, you have to perform all of a collection of sub-tasks or stages. If the number of ways of doing the sub-tasks are a, b, c, and so on, then the total number of ways to perform the task is a bc.Thisiscalledthemultiplicative counting principle.

Example:

Rajs favourite restaurant has a fixed menu from which he can choose an appetizer from three choices, a main course from three choices, and a dessert from two choices. How many different ways can Raj order his meal?

Solution:

In order for Raj to order his meal, he needs to order an appetizer, a main course, and a dessert.

The number of ways to accomplish the first task (ordering an appetizer), a =3

The number of ways to accomplish the second task (ordering a main course), b =3

The number of ways to accomplish the third task (ordering dessert), c =2

You can use the multiplicative counting principle to find the number of different ways that Raj can order his meal, which is:

a bc=332=18

Rajcanorderhismealin18differentways.



S u p p o r t Q u e s t i o n

Do not send your answers in for evaluation.

1. Lisa decides to fly from Windsor to Vancouver and return by train. For her flight, she can choose from three airlines: High Flights, ZoomAir, and Fly by Canada. For her trip back, she can travel by train with either QuickRail or ZoomRail. Draw a tree diagram and find the number of different ways in which Lisa can make the round trip to Vancouver.

There are Suggested Answers to Support Questions at the end of this unit.



Permutations In this section, you are concerned with problems where the order of selection is important. An arrangement of objects where order matters is called a permutation. (A selection of objects where order doesnt matter is called a combination; you will learn about combinations in the next lesson.)

In the introduction, you saw examples of the kinds of problems that organized counting tries to solve. Here are those examples again:

Ifatrackteamconsistsof12members,howmanywayscanthecoachchoosefourofthe members to fill the four different positions in a relay race?

Ifanicecreamstoresells31differentflavoursoficecream,howmanywayscanyouorder a cone that contains three scoops of ice cream?

Insomelotterygamesyouchoosesixnumbersfromthenumbers1through49.Howmany different possible lottery tickets are there?

One thing you might have noticed about the above examples is that order is important in some of the questions, but not in others. In the relay example, if the coach puts Pedro in the third position and Jason in the fourth position, thats different from putting Jason in third and Pedro in fourth, because you usually put the fastest person on your team in the fourth position. Because order is important in this case, this is a permutation problem. In contrast, in the lottery ticket example, the order in which the numbers on the ticket are chosen (and the order in which the winning numbers are drawn) is not important. Because order is not important, this is a combination problem. You will learn how to solve combination problems in the next lesson. The ice cream example could be either a combination problem or a permutation problem, depending on whether you cared if a particular kind of ice cream was on the top, in the middle, or at the bottom. If this sequence didnt matter, it would be a combination problem.

Now,supposeyoureplacethenumbers12,31,or49intheaboveexampleswithsmallernumbers. Then it might be possible to answer the questions simply by writing out all of the possibilities. You will look at some examples with smaller numbers, where you can do just that. These examples with smaller numbers will be used to illustrate general methods that you can use to solve problems with bigger numbers.

Heres an example using small numbers to illustrate how you might go about counting permutations.

Example:

A small ice cream store has four different flavours of ice cream for sale: Apricot, Blueberry, Coffee, and Dutch Chocolate. Ivanka wants to buy a cone with two scoops of ice cream. Shes very particular about her ice cream, and considers a scoop of Apricot on top of a scoop of Coffee to be different from a scoop of Coffee on top of a scoop of Apricot! She also doesnt mind having two scoops of the same flavour. How many possibilities are there for Ivankas ice cream cone?



Solution:

Ivankas cone is going to be built in two stages. First, a scoop of one of the four flavours of ice cream (A, B, C, or D) is placed on the cone. This can be done in four different ways, one for each flavour. Next, the second scoop is placed on top of the first scoop. This can also be done in four different ways. The numbers here are small enough to make it easy for Ivanka to have an exhaustive list of all of the different possibilities:

First scoop is A First scoop is B First scoop is C First scoop is DAA BA CA DAAB BB CB DBAC BC CC DCAD BD CD DD

Notice that there are four ways to choose the first scoop and then four ways to choose the secondscoop.Whenyoucancountupallofthepossibleways,youget16.

What if Ivanka decides she doesnt want two scoops of the same flavour? For this new problem, you can also make an exhaustive list of all of the possibilities:

First scoop is A First scoop is B First scoop is C First scoop is DAB BA CA DAAC BC CB DBAD BD CD DC

Whenyoucancountallofthepossibleoutcomes,youget12.

Here is another way of arriving at the answer in the ice cream example. Consider each scoop as a separate choice. The first choice represents the first scoop of ice cream placed on the cone, and the second choice represents the second scoop, which is placed on top of the first.

If Ivanka doesnt mind having two scoops of the same flavour, then there are four different ways for the first choice to occur and four different ways for the second choice to occur. The total number of possibilities in that case is:

44=16

Does this approach look familiar? What principle do you think is being used here? It is the multiplicative counting principle.

Now suppose that Ivanka doesnt want two scoops of the same flavour. There are still four different ways for the first choice to be made, but once the first choice is made, there are now only three different ways for the second choice to occur. The total number of possibilities in this case is:

43=12

Heres another more complex example using the multiplicative counting principle.



Example:

Hannah wants to choose a personal identification number (PIN) for her bank card. It has tobefourdigitslong.Eachdigitcanbeanynumberfrom0to9,butHannahdoesntwantto repeat any digits. How many different possible PINs can Hannah create?

Solution:

To answer this using the multiplicative counting principle, start by considering the number of separate choices you need to make to solve this problem. Since the PIN must be four digits long, you need to make four separate choices. The first digit can be filled withanyofthenumbersfrom0to9,sothereare10possibilities.Then,oncethefirstdigit has been chosen, there are nine possibilities for the second digit (remember that the second digit can be any number except whatever was chosen for the first digit). After that, there are eight possibilities for the third digit (that is, it can be any number except the ones chosen for the first two digits). Finally, there are seven possibilities left for the last digit. The total number of possible personal identification numbers that Hannah can create is:

10987=5040

Here is a more complicated example with more conditions on the desired outcome.

Example:

Howmanythree-digitnumberscanbeformedfromthefivenumbers{1,2,3,4,5},ifeachthree-digit number must be odd?

Solution:

Again, begin with deciding how many separate choices need to be made with the different numbers. Since you want a three-digit number, you need to make three separate choices. You must also account for the fact that the number must be odd, so the last digit can only bea1,3,or5.Thus,thelastdigit(ontheright)maybefilledinonlythreedifferentways:

____3

The first digit of the number can be any of the four remaining numbers, so there are four possibilities for it. Finally, there are three possibilities for the second digit. So, you get:

433=36

Therefore, the total number of odd three-digit numbers that can be formed from the numbers{1,2,3,4,5}is36.

Not all problems can be solved in just one step, like the ones you have just looked at. Here is an example of a problem that requires a two-part solution.

Example:

Howmanyfour-digitnumberscanbeformedfromtheeightnumbers{0,1,2,3,4,5,6,7},iftheresultingfour-digitnumberscanonlybeevennumbers?



Solution:

Sincethenumbermustbeeven,thelastdigitontherightmustbeoneof{0,2,4,6}.Thisproblem is a bit more challenging than the odd-number case you have just done, because the number of options for the first digit will depend on whether the last digit is a zero or not. This means that there are potentially two simultaneous restrictions on the location for zeroit cant be in the first position by definition, and it may have to be in the last position, to get an even number.

Its easiest to solve this problem in two parts. First, solve for those even numbers with a zero in the last digit and second, for those even numbers that do not end in zero.

Even numbers with 0 on the right:

Zero is the rightmost number, so there is only one way to choose the rightmost number. Starting from the left now, you can choose the first number in seven different ways, the second number in six different ways, and the third number in five different ways. The number of even numbers that can be formed from the above set with 0 on the right is:

7651=210

Even numbers without the 0 on the right:

Inthiscase,youhavethreechoicesfortherightmostdigit(thatis,2,4,or6).Thefirstdigit can be chosen from six numbers, because it cant be the even number chosen on the right or zero. However, you still have six choices for the second digit, because it can include zero. Finally, you have five choices for the third digit and you already know that you have three choices for the fourth digit. So, the number of even numbers that can be formed without a 0 is:

6653=540

Therefore,thetotalnumberofevennumbersis210+540=750.

S u p p o r t Q u e s t i o n s


2. A store sells six different kinds of toothpaste and five different styles of toothbrush. How many ways can a customer choose one tube of toothpaste, and then one toothbrush?

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3. Howmanyfour-digitnumberscanbeformedfromthenumbers{1,2,3,4,5,6},ifyou only want the result to be even numbers?

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Ordered N-tuples Any sequence of two elements where order matters (for example, where [A,B] is considered to be different from [B,A]), is called an ordered pair. Similarly, a sequence of three elements where order is important can be called an ordered triple; an ordered sequence of four elements can be called an ordered quadruple; and so on. The general way of naming these is to call them ordered N-tuples.

Using this new terminology, the example with Hannahs personal identification number can be rephrased as: How many ordered quadruples can be formed using the digits 0 through9,ifrepetitionisnotallowed?

Problems involving ordered N-tuples can also be solved using the multiplicative counting principle.

Example:

Howmanyorderedtriplescanbeformedfromasetcontaining17objectsifrepetitionisallowed? What if repetition is not allowed?

Solution:

Ifrepetitionisallowed,thenthefirstobjectcanbechosenin17differentways.Thesecondobjectcanalsobechosenin17differentways,andthethirdobjectcanbechosenin17differentwaysaswell.Therefore,usingthemultiplicativecountingprinciple,thetotal number of possible ordered triples is:

171717=4913

However,ifrepetitionisnotallowed,thentherearestill17differentwaystochoosethefirstobject,butonly16waystochoosethesecondobject,andonly15waystochoosethethird object. In that case, the total number of possible ordered triples is:

171615=4080

Here is an example that involves the arrangement of the letters of a word.



Example:

How many different ways can you arrange the letters of the word PHONE if:

a) there are no restrictions?

Solution:

Thinkofthequestionas,Howmanydifferentordered5-tuples(quintuples)arepossiblefromthesetofthefiveletters{P,H,O,N,E}?Therearefivedifferentwaysto choose the first letter in the arrangement, four different ways to choose the second letter, three different ways to choose the third letter, two different ways to choose the fourth letter, and finally, one way to choose the last letter. Thus, the total number of arrangements of the letters of the word PHONE is:

54321=120

b) the first letter must be a vowel?

Solution:

There are two vowels in the word PHONE: O and E. Since the first letter has to be a vowel, there are only two ways to choose the first letter in the arrangement. There are four possibilities left for the second position, three for the third position, two for the second position, and one for the last position. Thus, the total number of arrangements of the letters where the first must be a vowel is:

24321=48

c) the first letter is a consonant and the last letter is a vowel?

Solution:

There are three consonants and two vowels in the word PHONE. To solve this problem, start by finding the possible ways to fill the first letter and the last one. There are three ways to fill the first letter (with a consonant) and two ways to fill the last letter (with a vowel). After filling the first and last letters, you are then left with three choices for the second letter, two for the third, and one for the fourth. Therefore, the total number of arrangements of the letters where the first letter is a consonant and the last letter is a vowel is:

33212=36





4. Howmanyorderedtriplescanbeformed,usingtheletters{A,B,C,D,E,F,G,H}if

a) repetition is allowed?

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b) the second letter must be different from the first letter?

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5. How many different ways can the letters of the word ARTICHOKE be arranged if

a) there are no restrictions?

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b) the first letter must be a vowel and the second letter must be a consonant?

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c) the first letter must be a vowel and the eighth letter must be a consonant?

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6. How many licence plates can be made if

a) all of the licence plates have three letters followed by three numbers?

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b) no letter can be repeated on a licence plate, but numbers can?

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7. How many four-digit numbers are possible?

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8. How many seven-digit telephone numbers are possible? (Note that telephone numbers cannot start with 0.)

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Factorial Notation and Permutations In this section, you will learn how to simplify the work required in finding the number of ways to arrange and choose objects from a group. You have already learned that when the numbers are small, you can find the answer either by listing all of the possible outcomes or by using the multiplicative counting principle. But what about situations where the numbers get too big to work out easily using these methods? Is there an easier way to do this using a general formula? Fortunately there is, but first you need to learn about factorials.

Factorial Notation You have already seen situations where you are looking for the number of ways in which you can arrange a given number of objects. Heres a similar example.

Example:

Six people are lining up in a row to have their picture taken. How many different ways can they line up?

Solution:

If you solve this problem using the multiplicative counting principle, you will need to make six choices, one for each position in the row. The first position in the row can be filled in six ways (by any of the six people). Once the first choice is made, the next position can be filled in five ways (by any of the five remaining people). The next position can be filled in four ways, and so on. Using the multiplicative counting principle, the total number of ways that six people can line up is:

654321=720

In other words, the number of possible ways in which to arrange six distinct objects inalineistheproductofallpositiveintegersfrom1to6.Similarly,foranyvalueofn, the number of different ways to arrange n different objects in a line is the product of all positiveintegersfrom1ton.

Theproductofallpositiveintegersfrom1ton has a special name. It is called the factorial of n, or n factorial, and is written n! (n followed by an exclamation mark). So, for example, you have:

1!=1

2!=21=2

3!=321=6

4!=4321=24

5!=54321=120

6!=654321=720



It is important to note that factorials are only defined for positive integers. They are given a special name and notation because they are often used in many counting problems.

Itisalsoveryimportanttonotethat0!=1.Thismightseemstrange,butifyouthinkabout the number of ways to arrange nothing, there is only one way to do it.

Here are some examples that will show you how to work with, and manipulate factorials.

Example:

Simplify 19!16!

Solution:

19!16!

=191817 1615 ...1

1615 ...1= 191817 = 5814

Noticehowyoudonthavetofindtheactualvaluesof19!and16!.Sincethenumeratoristheproductofallintegersfrom1to19,andthedenominatoristheproductofallintegersfrom1to16,thenyoucandividealltheliketerms,makingtheequationalotsimpler.

Example:

Simplify:

a)

8!(8 3)!

Solution:

8!

(8 3)!=

8!5!

=8 7 6 5 4 ... 1

5 4 ... 1= 8 7 6 = 336

Noticethatyoucaneliminatethecommonterms541inthenumeratoranddenominator, making the equation a lot simpler.

b)

n!(n 4)!

Solution:

Again, you can eliminate the common terms in the numerator and denominator, which, in this case, are (n4)(n3)1.

c) (n +1)!(n 2)!

Solution:

(n + 1)!(n 2)!

=(n + 1) n (n 1) (n 2) (n 3) ... 1

(n 2) (n 3) ... 1= (n + 1) n (n 1)

n!(n 4)!

=n(n1)(n 2)(n3)(n 4)(n5) ...1

(n 4)(n5) ...1= n(n1)(n 2)(n3) n(n1)(n2)(n3)



The following examples show how to solve equations involving factorials.

Example:

Solve for n in: (n +1)!

n!= 16

Solution:

Simplify the left-hand side. Start by expanding the factorials in the numerator and denominator and look for common terms:

(n +1)!

n!=

(n +1) (n) (n 1) (n 2) (n 3) (n 4) ...1(n) (n 1) (n 2) (n 3) (n 4) ...1

Notice that you can eliminate the common terms (n)(n 1)(n 2)(n 3) (n 4). . . 1inthenumeratoranddenominator.Thissimplifiestheleft-handside of the equation to:

n+1

Hence, n+1=16,andson =15.

Example:

Solve for n:

n!(n 2)!

= 42

Solution:

You simplify the left-hand side by eliminating the common terms in the numerator and denominator:

n!

(n 2)!=

n (n 1) (n 2) ...1(n 2) ...1

= n(n 1)n(n1)

So, the equation becomes n(n 1)=42.Thiscanbeexpandedintothefollowingquadratic equation:

n2 n42=0

To solve this equation, use the quadratic formula. Recall that the quadratic formula gives the solutions to quadratic equations of the form: ax2 + bx + c = 0.

The quadratic formula is:

x = b b

2 4ac

2a

In this example, a=1,b=1,c=42andx = n.



So, the quadratic formula becomes:

n = (1) (1)2 4(1)(42)

2

=1 169

2

=1 13

2

The solutions to n2 n42=0are n = 14

2= 7 or

n = 12

2= 6.

Since n must be a positive integer, then n=7. Since n is a positive integer, n2 n42=0canbesolvedbyfactoring(n7)(n+6)=0n=7orn=6

n=7



9. Simplify:

a) 5!3!

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b) (n +1)!

n! _______________________________________________________________

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c) (n + 3)!(n 1)!

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10. Solve for n:

(n +1)!(n 1)!

= 132

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Solving Permutations using Factorial Notation Now you are going to study situations in which you are looking at the number of arrangements of r objects that you can get out of n objects. The problems you worked on earlier involving Hannahs choice of personal identification number (PIN) are an example of this type of arrangement.

Example:

Recall earlier that Hannah wanted to choose a PIN number for her bank card. It had to be fourdigitslong.Eachdigitcouldbeanynumberfrom0to9,butHannahdidntwanttorepeat any digits. How many different possible PIN numbers are there?

Solution:

Inthisexample,youarelookingatthenumberoffour-itemarrangementsoutof10.Recall,earlierinthelesson,thattheanswertothiswas10987arrangements.

Using factorial notation, this can be written as 10!6!

.

In general, an arrangement of r objects chosen from n things is called a permutation of n things taken r at a time. It is denoted by the symbol P(n,r) or nPr . The formula to calculate a permutation is:

P(n,r) = n!

(n r)!

For example, the personal identification number scenarios would be written as:

P(10, 4) = 10!

6!=

10 9 8 7 6 5 4 3 2 16 5 4 3 2 1

= 10 9 8 7 = 5040

Example:

Use factorial notation to find the number of ordered triplets that can be formed from a set of eight elements.

This question is asking for the number of arrangements of three elements that can be chosen from the eight elements. Or phrased another way, find the number of permutations of three elements from a set of eight. The answer, using factorial notation, is:

P(8, 3) = 8!

(8 3)!=

8!5!=

8 7 65 4 3 215 4 3 21

= 8 7 6 = 336



Here are some additional examples in shorter form:

P(17,12)=17161514131211109876

P(8,3)=876

P(21,5)=2120191817

Hint: Notice that in the above examples, the final answer can also be found simply by multiplying the first r terms of n!. So, the quick way to find P(9,4)isjusttomultiplythefirstfourtermsof9!(thatis,9876),whichequals3024.Usingthesamelogic,youcan see that the answer to P(5,2)wouldbe54=20andthatP(14,3)=141312= 2184.

More Complex Permutation Problems

What if there are some restrictions on the arrangement of objects in a permutation? For example, what if the problem asked, In how many ways can five different books be arranged on a shelf, if two specific books must always remain together?

Since two of the books must be kept together, you can think of them as one book. Therefore, there are really only four different books to be arranged in P(4,4)ways.Thetwo books that must remain together can be arranged in P(2,2)ways(suchasTreasure Island and Hamlet, or Hamlet and Treasure Island).

Thus, the number of arrangements is:

P(4,4) P(2,2) =(4321)(21)=48

YouwilllookatmorecomplicatedscenariosofthistypeinLesson8.



11. Find the value of each of the following using the formula P(n,r).

a) P(7,2)

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b) P(4,1)

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c) P(12,2)

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d) P(11,4)

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12. Given the set A ={2,3,4,5,7}:

a) How many ordered pairs can be formed from Set A, if repetition is allowed?

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b) How many ordered pairs can be formed from Set A, if repetition is not allowed?

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13. How many ways can a president, a vice-president, and a treasurer be chosen from 10candidates?

a) Solve this question using the permutation formula.

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b) Check your answer using the multiplicative counting principle.

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14. How many four-letter words can be formed from the word HORSE, if no letter can be used twice?

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15. Given the set L={R,S,T,U,V,W}:

a) How many five-letter words can be formed from L without repeating any letters in the words?

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b) How many six-letter words can be formed from the letters in part (a) if the letters T and U must always be together?

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Key ConceptsCombinatorics: the branch of mathematics that deals with counting the number of ways in which something can be done

Multiplicative counting principle: a principle stating that if the numbers of ways of doing sub-tasks are a, b, c, and so on, then the total number of ways to perform the task is a bc

Permutation: an arrangement of objects where order is important

Ordered pair, ordered triple, ordered quadruple, ordered N-tuples, and so on: a sequence of two (in the case of a pair) or three (in the case of a triple) or more elements (in the case of N-tuples), where order is important

Factorial: theproductofallpositiveintegersfrom1ton. It is called the factorial of n, or n factorial, or n!.



K e y Q u e s t i o n s

Save your answers to the Key Questions. When you have completed the unit, submit them to ILC for marking.

Total: 25 marks

19. Simplify: (7 marks total)

a) 23!20!

(1 mark)

b)

(n + 2)!(n + 2)n!

(2 marks)

c) Solve for n: (4 marks)

(n + 2)!

(n)!= 110

20. How many different five-digit numbers can be formed from the set of seven numbers{1,2,3,4,5,6,7}if

a) repetition is allowed?

b) repetition is not allowed?

c) repetition is allowed and the third digit must be different from the second digit?

Show your calculations. (6 marks: 2 marks for each part)

21. Determine the number of ways of arranging the letters of the word MUSCLE if

a) there are no restrictions. (2 marks)

b) the third and fourth letters must both be consonants. (3 marks)

Show your calculations. (5 marks total)

22. How many possible Canadian postal codes are there? (Recall that our postal codes have the following form: letter number letter, number letter number.)

Show your calculations. (2 marks)

23. Howmanyfive-digitnumberscanbeformedfromthesetofninenumbers{0,1,2,3,4,5,6,7,8}ifnonumberisrepeatedandnonumberstartswithazero,and

a) there are no other restrictions? (2 marks)

b) the result must be an odd number? (3 marks)

Show your calculations. (5 marks total)



Now go on to Lesson 7. Do not submit your coursework to ILC until you have completed Unit 2 (Lessons 6 to 10).


unit 2 - lesson 6

Documents

counting problems

principle combinatorics

organized counting techniques

grocery store

number of possible outcomes

important principle

different routes

farmers market