VCEcoveragemathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 11...The adhesives are PVA glue, Liquid Nails and Bondcrete. Fasteners that can be used are nails, screws, rivets and

VCEVCEcocovverageerageArea of studyUnit 2 • Probability

In thisIn this chachapterpter11A Addition and multiplication

principle11B Permutations11C Factorials11D Permutations using nPr

11E Permutations involving restrictions

11F Arrangements in a circle11G Combinations using nCr

11H Applications to probability

11

Combinatorics

476

M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Introduction

Consider how many ways two different letters can be listed from the letters C, A and Tif the order in which the letters are to be listed is not taken into account. We would writeCA, CT and AT. If the order of listing the two letters is taken into consideration, therewill be 6 different possibilities, namely, CA, AC, CT, TC, AT and TA. In this chapter weintroduce some methods that will enable us to effectively determine the number of possibleways objects can be ordered according to given conditions, without necessarily having tolist them. Combinatorial theory is widely applied in areas such as computer system design,genetics, statistics and probability, where arrangements are of particular importance.

The addition principle

To reach the top of the hill, Jack and Jill can use public transport (tram or bus) or privatetransport (hire-car, taxi or motorbike). In how many ways can Jack and Jill go up thehill if only one form of transport is to be used for the entire trip?

Since the modes of transport are mutually exclusive (that is, two forms of transportcannot be used at the same time), there are 2

+

3

=

5 different ways of travel.This straightforward method of summing is the

addition principle

that can be stated as:If two operations can be performed in A or B ways respectively, then both operationscan be performed together in A

+

B ways.

A particular mathematics problem can be solved in 2 ways using analytical methods, in 4 ways using approximation techniques and in 3 ways by trial and error strategies. In how many ways can the problem be solved?THINK WRITE

List the given information. Analytical 2Approximation 4Trial and error 3

Addition is involved because the three methods of solving the problem are mutually exclusive.Sum the number of ways for all the methods. Total number of ways is 2 + 4 + 3 = 9.

1

2

3

1WORKEDExample

As part of a word game, Gary must correctly find 3 secret words by choosing one letter of the alphabet at a time. He has established that the 3 words are of the form ST_P, THA_ and N_W. If it is known that the 3 missing letters are a vowel, a consonant and a vowel respectively, find the maximum number of guesses Gary would need to find all 3 words.THINK WRITE

List the given information. Letter 1 Letter 2 Letter 3vowel consonant vowel

Consider how many vowels and consonants there are in the English alphabet (5, 21).

5 21 5

Sum the maximum number of guesses possible for the 3 words.

The maximum possible number of guesses is 5 + 21 + 5 = 31.

1

2

3

2WORKEDExample

C h a p t e r 1 1 C o m b i n a t o r i c s

477

Multiplication principle

Suppose 4 colours are available to spray-paint 3 different cars. Let O

1

be the first oper-ation — selecting a car — and let O

2

be the second operation — picking a paint colour.Also let C

1

, C

2

, C

3

and P

1

, P

2

, P

3

, P

4

denote, respectively, the cars and the availablecolours. Then the 12 different ways in which the job can be undertaken are {C

1

P

1

, C

1

P

2

,C

1

P

3

, C

1

P

4

, C

2

P

1

, C

2

P

2

, C

2

P

3

, C

2

P

4

, C

3

P

1

, C

3

P

2

, C

3

P

3

, C

3

P

4

}. Since the choice of aparticular car is independent of the colour selected, the total number of possibilities canbe obtained by multiplying together the number of choices available from the twooperations. That is, there are 3

×

4

= 12 different ways possible. The tree diagram belowshows the different outcomes.

The product of the number of outcomes from each operation readily provides thetotal number of possible outcomes of the operations performed sequentially.This method is the basis of the multiplication principle that states:If two operations can be performed in A and B ways respectively, then both operationscan be performed in succession in A × B ways.

We can also represent the sequence of operations of the above example using boxednumbers as follows.

Each box contains the number of possible outcomes associated with the particularoperation.

Skirts Jumpers Shirts

6 5 8 = 240

P1P2P3P4

P1P2P3P4

P1P2P3P4

C1

C2

C3

Operation 1 Operation 2O1 O2

Juanita has to choose an outfit to wear to a party. She has 6 skirts, 5 jumpers and 8 shirts to choose from. If any combination of these items will be acceptable attire, in how many styles of dress can Juanita attend the party?

THINK WRITE

There are three operations.Choose a skirt, a jumper and a shirt. There are 6 skirts, 5 jumpers, 8 shirtsUse the multiplication principle. Total number of ways is 6 × 5 × 8 = 240.

1

2

3

3WORKEDExample

478 M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

From a cafeteria 4-course lunch menu, I can choose 3 varieties of soup, 5 types of seafood, 4 kinds of side dish and 2 types of salad.a How many different dishes are offered?b How many different lunches can be ordered if one dish from each course is selected?c How many different types of dish are possible if soup and seafood must be included

with each order?

THINK WRITE

a Write the number of dishes for each course and use the addition rule.

a The 4-course menu offers 3 + 5 + 4 + 2 = 14 different dishes.

b There are 3 soups, 5 seafoods, 4 side dishes and 2 salads. Use the multiplication rule.

b

Number of different lunches is 120.

c Consider the possible orders containing soup and seafood:

c

soup and seafood only

soup and seafood and a side dish only

soup and seafood and a salad only

soup and seafood and a side dish and a salad.

Calculate the number of dishes possible for each order.

Use the addition rule to find the total.

Number of different types of dish possible= 15 + 60 + 30 + 120= 225

Soup Seafood Side dish Salad

3 5 4 2 = 120

1Soup Seafood

3 5 3 × 5 = 15

Soup Seafood Side dish

3 5 4 3 × 5 × 4 = 60

Soup Seafood Salad

3 5 2 3 × 5 × 2 = 30

Soup Seafood Side dish Salad

3 5 4 2 3 × 5 × 4 × 2 = 120

2

3

4WORKEDExample

remember1. Addition principle: If two operations can be performed in A or B ways

respectively, then both operations can be performed together in A + B ways.2. Multiplication principle: If two operations can be performed in A and B ways

respectively, then both operations can be performed in succession in A × B ways.

remember

C h a p t e r 1 1 C o m b i n a t o r i c s 479

Addition and multiplication principle

1 Juicy Chickens offers 10 varieties of roast chicken dish, 6 types of fried chicken and5 types of chicken pie. How many different chicken meals are sold by JuicyChickens?

2 Freda Frog eats 2 varieties of fly on the first day, 5 varieties on the second day, 9varieties on the third day and 14 varieties on the fourth day. Assuming Freda willnever consume 2 of the same variety of fly and that her daily eating habits followthis definite pattern, find how many flies she will eat altogether in a week.

3 A suburban mall consists of five shops: Teen Fashion, Harry’s Takeaway, Video &Games Arcade, Toy Palace and Byte Computers. On a busy weekend, 11 peoplewent into Teen Fashion, 27 bought food from Harry’s Takeaway and 59 people pat-ronised the Toy Palace. Each person visited only one store. How many customersdid Teen Fashion, Harry’s Takeaway and Toy Palace have altogether?

4Two pieces of timber can be held together using adhesives, fasteners or clamps.The adhesives are PVA glue, Liquid Nails and Bondcrete. Fasteners that can beused are nails, screws, rivets and bolts. There are two different types of clampsavailable: SureGrip and Hold-tite. The number of ways two pieces of timber can bejoined is:

5 From a pack of playing cards are drawn the queen of spades, king of clubs andqueen of clubs. In how many ways can another card from the deck be drawn sothat there will be three queens or two kings?

6Hassan is guessing the ages of Julian and Dennis. He knows that Julian’s age is an evennumber less than 20 and that Dennis’ age is an odd number between 18 and 30. Themaximum number of guesses Hassan can make before correctly guessing the two ages is:

7 Jack and Diane are preparing for their wedding. They will decide on one of 3churches, one of 5 available reception centres and one of 10 holiday destinations.How many combinations of church, reception centre and holiday are possible?

8 Alana lives in Melbourne and intends to go to Sydney via Canberra. She will get toCanberra by bus, continue on to Sydney by hire-car and return home by air. If 4bus lines are available for the outward journey to Canberra, 6 car rental agenciescan be used to get from Canberra to Sydney and 3 airlines are available for thereturn trip, determine how many different ways Alana can make the trip to Sydneyand back.

A 2 B 24 C 3 D 4 E 9

A 12 B 9 C 19 D 15 E 25

11AWORKEDExample

1

mmultiple choiceultiple choice

WORKEDExample

2


WORKEDExample

3


9 At Burpies restaurant the ‘special meal’ consists of a choice of one of 2 entrèes, oneof 3 main meat dishes and one of 4 kinds of dessert. For a ‘surprise feast’ at Belchiesrestaurant you can have one of 5 different entrèes, select from 4 main meals anddecide which one of 3 kinds of dessert to order.a How many different combinations of dishes are possible in a ‘special meal’

consisting of an entrèe, a meat dish and a dessert?b Find how many different combinations of dishes are available to a customer who

visits both places and orders a ‘special meal’ and a ‘surprise feast’. (Assume thatthe customer must have an entrèe, main meal and dessert for the ‘surprise feast’.)

10On a dentist’s waiting room table are 3 piles of reading matter. The first pile consistsof 6 different copies of News, the second pile has 5 different issues of Geographic andthe third pile comprises 10 different Woman’s World magazines. A nervous patientrandomly chooses one item of reading from each pile. The number of ways ofchoosing the 3 items is:

11 A school offers English, Maths, Language and Science as part of the curriculum.Janice must do at least one of these subjects. a List the different ways Janice can select at least one subject.b In how many ways can this be done?

12A Whoppa pizza base is madeusing one of 2 types of cheese andone of 2 toppings. Up to 3 addi-tional toppings are available atextra cost. The number of dif-ferent Whoppa pizzas that can bemade containing at least oneadditional topping is:

13 To get to school, Erin can walk,take the train or catch the bus.After school she can either walkor catch the bus to get back home.a List the different combinations of travel for Erin to get to school and to return to

her home.b Show the different travel methods as a tree diagram.

14 A hot-dog consists of a sausage in a bun with sauce. Onion, tomato, pineapple andcheese are available as extras. How many different types of hot-dog can be made?

15 During a special morning recess, teachers had a choice of tea, orange juice, coffee, pies,cheese, salami, biscuits and cake. However, a teacher could sample only two kinds offood and one drink. How many different combinations of food and drink were possible?

A 21 B 30 C 216 D 19 E 300

A 12 B 16C 24 D 28E 20


WORKEDExample

4



PermutationsA permutation is the arrangement of objects in a specific order. Awarding a first andsecond prize to two people randomly selected from a studio game-show audience ordetermining the number of ways a group of people can queue for tickets are exampleswhere the order of objects needs to be taken into account.

Consider now how many ways two letters can be taken from the letters B, L, U andE and then arranged.

The order of the letters is taken into account and repetition of letters is not allowed,so we have the 12 possible arrangements shown below:

BL, LB, BU, UB, BE, EB, LU, UL, LE, EL, UE, EUWe can obtain the same result using the multiplication principle. There are 4 choices

for the first letter because there are 4 letters available. Once the first letter has beenchosen there are 3 letters to choose from for the second letter.

Notice that the multiplication principle takes into account the order of the out-comes. That is, BL is not considered to be the same as LB, BU is not the same asUB and so on.

First letter Second letter

4 3 = 12

Josie picks up a Mathematics textbook, an English novel and a Biology notebook and places them on a shelf. Determine the number of arrangements and list them.

THINK WRITE

There are three positions to be filled on the shelf.

3 × 2 × 1 = 6 arrangementsThere are three choices of book for the first position on the shelf. This leaves two choices for the second position and one choice for the third position.Use the multiplication principle.Let M be the Mathematics textbook, E the English novel and B the Biology notebook.

The arrangements are MEB, MBE, BME, BEM, EMB, EBM.

13 2 1

2

34

5WORKEDExample

In how many ways can at least two letters be chosen from the word STAR if the order of the letters is taken into account and repetition of letters is not allowed?THINK WRITE

There are 3 mutually exclusive events: choose 2 letters from 4 letters, 3 letters from 4 letters, or 4 letters from 4 letters.

For the first event there are 4 choices for the first letter and 3 choices for the second letter since repetitions are not allowed.

2 letters = 12 ways

1

2 4 3

6WORKEDExample

Continued over page


THINK WRITE

For the second event there are 4 choices for the first letter, 3 choices for the second letter and 2 choices for the third letter.

3 letters = 24 ways

For the third event there are 4 choices for the first letter, 3 choices for the second letter, 2 choices for the third letter and 1 choice for the fourth letter.

4 letters = 24 ways

Use the addition rule to find the total number of possibilities.

Number of ways = 12 + 24 + 24Number of ways = 60

3 4 3 2

4 4 3 2 1

5

How many ways are there for 2 different prizes or 3 different prizes to be awarded to a group of 5 people if: a a person may receive more than one award?

b a person may not receive more than one award?THINK WRITE

a Use the multiplication principle to find the number of ways for 2 prizes to be awarded. Any one of the 5 people can receive the first prize and any one of the 5 people can receive the second prize because the same person may receive more than one prize.

a 2 prizes = 25

Use the multiplication principle to find the number of ways for 3 prizes to be awarded and remember that the same person may receive more than one prize.

3 prizes = 125

Use the addition rule to obtain the total. Number of ways to distribute 2 or 3 prizes= 25 + 125= 150

b Use the same method as above, but repetition is not allowed so that the number of people to choose from is reduced each time.

b 2 prizes = 20

3 prizes = 60Number of ways to distribute 2 or 3 prizes

= 20 + 60= 80

1

1st 2nd

5 5

2

1st 2nd 3rd

5 5 5

3

5 4

5 4 3

7WORKEDExample

remember1. A permutation is the arrangement of objects in a definite order.2. The multiplication principle is commonly used in calculating the number of

possible permutations.

remember


Permutations

1 A chef restocks her collection of spices by placing jars of pepper, nutmeg, ginger andmint on the shelf. In how many different ways can the 4 jars be placed in a straight line?

2 In how many ways can 6 students line up at the school canteen?

3 If there are 8 competitors in a race, in how many ways can the first three places beawarded?

4 To cancel an electronic alarm, a 5-digit number must be entered into the code box.What is the maximum number of tries I have to cancel the alarm?

5 Five items of mail are to be placed in 5 letterboxes. In how many ways can this bedone if no letterbox is to contain more than one item?

6 A history quiz consists of matching 8 countries with their capital cities. In how manyways can a contestant answer the quiz by randomly matching each country with acapital city?

7 How many ordered subsets consisting of two letters can be chosen from the wordSUPERBLY if:a a letter may be used more than once in each subsetb choosing the same letter more than once is not permitted.

8 In how many ways can at least two letters be chosen from the word MATHS if theorder of the letters is taken into account and repetition of letters is not allowed?

9 Calculate the number of ways that 3 or 4 prizes can be awarded to a group of 5 people if:a a member of the group is allowed to receive more than one prizeb a member of the group cannot receive more than one prize.

10 Decide in how many ways 2 or 3 letters can be selected from the vowels of thealphabet if a vowel can appear only once in each selection.

11 Determine how many numbers greater than 10 can be made using all of the digits4, 7, 2, 6 and 5, if each digit cannot be used more than once.

12 How many numbers greater than 100 and less than 10 000 may be formed using thedigits 2, 3, 4 and 5 if each digit may be used more than once?

13

The number of 3-digit and 4-digit numbers greater than 500 that can be formed usingthe digits 2, 6, 1, 5 and 3, if each digit can be used more than once in each selection, is:A 600 B 500 C 675 D 575 E 450

11BWORKEDExample

5

WORKEDExample

6

WORKEDExample

7



14

Juliana has saved her pocket money to buy up to 3 fashion magazines. If there are 5different magazines to choose from, the number of ways she can buy 1, 2 or 3magazines is:

15

The total number of 2-digit, 3-digit and 4-digit odd numbers that can be formed usingthe digits, 5, 3, 4 and 9, when a digit may be used more than once in a group, is:

16 How many arrangements are possible usingthe letters of the word DECAGON if theletters A, E and O must occupy the third, fifthand sixth positions respectively and the lettersremaining may be used more than once?

17 A school fundraising competition that costs5 cents per entry involves trying to correctlymatch 9 teachers with their baby photo-graphs. Wasim wants to be certain to win the$1000 first prize by trying all possible combi-nations. Decide how much money Wasimwill win or lose if he is to be the prize winner.

18 A version of the party game ‘musical chairs’ has the players march around a line ofchairs and scramble to sit on them when the music unexpectedly stops. At each stagethe number of players is one more than the number of chairs. The player who remainsstanding when the music stops is out of the game and one chair is then removed. Theplayer remaining sitting after all the other players have been eliminated is the winner.a If 12 players are taking part, how many different arrangements of seating are

possible during the:i first round? ii fourth round?

b The rules are changed so that 2 chairs are removed each time. If there are 9players and 7 chairs at the start of the game, how many seating arrangements arepossible for all the rounds?

A 90 B 80 C 25 D 70 E 85

A 78 B 85 C 252 D 68 E 75

Identification cardsA progressive school is using identification cards (ID cards) that consist of 3 letters selected from A to E inclusive followed by 3 digits chosen from 0 to 9 inclusive.1 How many different ID cards can be issued to students if a digit may be used

more than once but all 3 letters of each ID are different?2 New ID cards are issued to all students each year and the old cards discarded.

However, the old ID numbers are not used again. If, on average, the school’s population increases by 10% each year and was 2000 during the year when the ID cards were first used, how many years will elapse before cards with numbers already used will have to be issued?




FactorialsExpressions obtained by using the multiplication principle frequently contain theproduct of consecutive whole numbers. It is convenient to adopt a shorthand way ofrepresenting such expressions to assist with calculations and to effectively display theproperties associated with permutations and other types of order of objects. Particularlyuseful is to define n! to mean the product of n consecutive positive integers startingfrom n down to 1. That is:

n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1The symbol n! is called n factorial.For example, 4! = 4 × 3 × 2 × 1 = 24, 3! = 3 × 2 × 1 = 6, 2! = 2 × 1 = 2Alternatively, 4! = 4 × 3! = 24, 3! = 3 × 2! = 6, 2! = 2 × 1! = 2

Thus from the definition we have n! = n(n − 1)! or

If we substitute n = 1 we have:

or

This expression is true if 0! is taken to be equivalent to 1. So we define 0! = 1

n!n 1–( )!

------------------ n=

1!1 1–( )!

------------------ 1= 10( )!

--------- 1 since 1! = 1=

a Express 7! as a numeral. b Simplify 2 × 5! + (3 × 2)!

THINK WRITE

a Use the definition of n! with n = 7. a 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1

Multiply the numbers in the expression obtained.

7! = 5040

b Write the expression and simplify (3 × 2)!

b 2 × 5! + (3 × 2)!= 2 × 5! + 6!

Calculate 5! and 6! = 2 × 120 + 720

Evaluate. = 960

1

2

1

2

3

8WORKEDExample

Simplify

THINK WRITE

Divide the answer for 8! by the answer for 3! using a calculator.

=

= 6720

8!3!-----

8!3!----- 40 320

6----------------

9WORKEDExample


Notice that: n! = n × (n − 1)!= n × (n − 1) × (n − 2)!= n × (n − 1) × (n − 2) × (n − 3)! etc.

To enter a factorial, such as 8! on the TI–83,1. Enter the number (in this case 8) on the home screen.2. Press then select PRB, and then 4:!3. Press to evaluate.

Stirling’s formulaTo simplify calculation of n! for large values of n, eighteenth century Frenchmathematician Abraham De Moivre proposed the following formula, which henamed after Scottish mathematician James Stirling:

Sn = ,

where π and e are mathematical constants approximated by 3.142 and 1.278respectively.1 Set up a spreadsheet or use a graphics calculator (with 2nd [ENTRY] to repeat

similar calculations) to compare the values of n! and Sn for various values of n(or use the Maths Quest spreadsheet ‘Stirling’s formula’). Include a columnshowing percentage error.

2 Plot a graph showing the percentage error for various values of n.

Simplify

THINK WRITE

100! and 98! are too large to write in fully expanded form.

Express 100! with 98! as a factor.100! = 100 × 99 × (98 × . . . × 3 × 2 × 1) = 100 × 99 × 98!

=

Cancel 98! in the expression. = 100 × 99= 9900

100!98!

-----------

1

2100!98!

----------- 100 99 98!××98!

-----------------------------------

3

10WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Factorials

MATHENTER

2πnne---

n

EXCEL

Spreadsheet

Stirling’sformula


Factorials

1 Evaluate:

2 Evaluate:

3

The value of is:

4

The value of 4(4! − 3) + 2!(5! − 4!) is:

5 Simplify:

6 Simplify:

7 Simplify:

8

is equal to:

9 Evaluate each expression.

10 Simplify each expression.

a 4! b 9! c 12! d 3! + 2!e 5! − 4! f 7! − 6! − 2! g 6! − (1! + 2! + 3! + 4! + 5!)

a 4 × 3! − 4! b (4 + 2!) 3! + 5! c 5 × 6! − 6 × 5!d 7 × 7! − (8! − 7!) e 8! + 3 × 2! − 5! f 7 × 9! + 3 × 3! − 9 × 8!g (5! − 4!) + (8! − 7!) h 12! + 6! − 11! − 3 × 4!

A 5 B 4 C 7 D 24 E 8

A 250 B 235 C 284 D 276 E 290

a b c d e

a b c d e f

a b c d e

A 3250 B 1875 C 2840 D 1030 E 2352

a b c d e

a b c d e

remember1. The factorial of a positive whole number n is defined as:

n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 with 0! = 12. n! = n × (n − 1)!

= n × (n − 1) × (n − 2)! and so on.

remember

11CWORKEDExample

8a

WORKEDExample

8b


2! 3! 4!+ +2! 3!+

---------------------------


WORKEDExample

9 4!2!----- 5!

4!----- 7!

3!----- 6!

3!----- 3!

2!----- 1!

0!-----+

WORKEDExample

10 102!100!----------- 1000!

998!-------------- 4500!

4499!-------------- 250!

247!----------- 396!

393!----------- 25 000!

24 999!------------------

15!14!-------- 28!

26!-------- 55!

53!-------- 1000!

998!-------------- 63!

936 875–( )!------------------------------


12 37+( )!100 53–( )!

---------------------------

7!4!----- 4!

3!-----+ 9!

7!----- 32!

31!--------+ 13!

10!-------- 6!

3!-----– 80!

77!--------

10!6!

-------- 64!62!--------+– 8!

6!-----

12!11!-------- 78!

77!--------+–

WorkS

HEET 11.1

2!5!3!

---------- 6!2! 3!+---------------- 7! 5!–

5!---------------- 8! 4!+

2! 3!+---------------- 2!

3!-----× 18!4!

17!5!------------- 8!10!

9!9!-------------×


Permutations using nPrA permutation is an arrangement of objects in which order is important.Consider the letters A, B and C. There are 6 possible arrangements or permutations ofthese three letters:

ABC ACB BAC BCA CAB CBAWe could have determined that there are 6 possible arrangements without listing all

of them using the multiplication principle, where each box below represents a position(first letter, second letter, third letter):

= 6 ways

Note that we had 3 possibilities for the first letter, but having placed it, were left with2 possibilities for the second letter, and in turn just 1 possibility for the third.But what if we had 10 different letters, and wished to arrange them taking only 3 at a

time? Again, we could count the number of arrangements as follows:

= 720 ways

If we wish to know only how many ways the 10 letters can be arranged 3 at a time,this is certainly less painful than trying to list all 720 arrangements!

We can express the above calculation using factorials as follows:

10 × 9 × 8 =

=

=

Following on from this, we can generalise a formula for the number of arrangements(permutations) of n objects, taking r at a time, which we denote by nPr:

nPr = where n and r are natural (counting) numbers, and r ≤ n.

Another way of thinking of nPr is as n! expanded to r ‘places’.nPr =

r values multiplied togetherIn the preceding example which involved arranging 10 (n = 10) objects (letters)

taking 3 (r = 3) at a time, we can verify that (n − r + 1) = (10 − 3 + 1) = 7, which wasindeed the last value in the chain of multiplied numbers.

Special cases1. If r = 0, then nPr = nP0

== 1

1st letter

2nd letter

3rd letter

3 2 1

1st letter

2nd letter

3rd letter

10 9 8

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

----------------------------------------------------------------------------------------- / / / / / / / / / / / /

10!7!

--------

10!10 3–( )!

----------------------

n!n r–( )!

------------------

n n 1–( ) n 2–( ) . . . n r– 1+( )

n!n!-----

C h a p t e r 1 1 C o m b i n a t o r i c s 489This implies that there is one way of selecting zero objects from n objects.

2. If r = n, then nPr = nPn

== n!

There are n! ways of arranging n objects taken from n objects.

n!0!-----

Calculate 7P3.THINK WRITE

Evaluate using the definition nPr.7P3 =

=

=

= 210

7!7 3–( )!

-------------------

7!4!-----

504024

------------

11WORKEDExample

Only the 4 fastest cars in a car rally of 20 competitors will compete in the world championships. How many different arrangements of the 4 fastest rally cars are possible?THINK WRITE

We want the number of permutations when 4 objects are selected from 20 objects.Use nPr for r = 4, n = 20. Number of arrangements

= 20P4Use the calculator function nPr or the multiplication principle.

= 116 280

1

2

3

12WORKEDExample

How many numbers greater than 1000 can be formed using the digits 4, 5, 6, 7, 8 and 9 if a digit cannot be used more than once?

THINK WRITE

Each 4-digit, 5-digit or 6-digit number formed will be greater than 1000.Find the number of ways the required number of digits can be chosen from the 6 digits given.

6P4 + 6P5 + 6P6

Add the 3 answers.(‘or’ situation)

= 360 + 720 + 720= 1800There are 1800 numbers greater than 1000 that can be formed.

1

2

3

13WORKEDExample


To calculate nPr using the TI–83 on the home screen, type in the value of n first, thendo the following:

1. Press PRB 2:nPr.

2. Type the value of r, and press .The screen dump below is for finding 10P3.

A captain and vice-captain are to be chosen from a group consisting of 10 cricket players. From the remaining 8 players, 3 will be selected to be the wicket keeper, spin bowler and fast bowler. Calculate how many different ways the 5 positions can be allocated.

THINK WRITE

Find the number of ways in which 2 objects (captain/vice-captain) can be chosen from 10 objects (10 cricket players).

Find the number of ways in which 3 objects (wicket keeper/spin bowler/fast bowler) can be chosen from 8 objects (8 cricket players).

Number of different ways= 10P2 × 8P3= 90 × 336

Multiply the two results.(‘and’ situation)

= 30 240

1

2

3

14WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Permutations

MATH

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remember1. A permutation is an arrangement of objects in which order is important.2. The number of arrangements of n objects, taken r at a time, is given by nPr.

remember


Permutations using nPr

1 Evaluate:

2 A committee comprising a president, vice-president, secretary and treasurer is to beselected from a group of 25 people. How many different committees are possible?

3 In how many ways can a first and second prize be given to 5 lottery winners?

4 How many numbers greater than 100 can be formed from the digits 2, 3, 5, 7, 9, if adigit cannot be used more than once?

5 John has a 5 cent coin, a 20 cent coin, a 50 cent coin and a $2 coin.a In how many ways can the coins be placed in a row?b In how many ways can 2 coins or 3 coins be chosen if the order is taken into

account?

6

A magic paint set contains seven ‘magic’ colours which when applied to paper pro-duce other colours. The colour obtained depends on the order in which the colours areapplied and at least two colours must be used. The number of different colours thatcan be produced is:

7 A captain and vice-captain are to be selected from a team of 18 footballers. From theremaining 16, four players will be selected to be the full-back, full-forward, centre-half back and centre-half forward. Calculate the number of ways the 6 positions canbe allocated.

8 The Southern Belle’s train crew consists of 2 drivers and 4 engineers. Each personperforms different tasks. The 2 drivers are chosen from 6 available drivers and the4 engineers from 10 engineers. How many permutations of the train’s crew arepossible?

9 Three students are to be chosen from a group of 8 students to fill the positions ofschool president, vice-president and treasurer. After these appointments are made,2 more students will be selected from the group to serve as secretary and assistantsecretary. Determine how many different committees are possible.

a 6P4 b 8P2

c 9P3 d 4P4

e 25P12 f 3P2

g 4P2 + 5P1 h 8P6 − 7P3

i 6P3 − 5P4 j 3P1 × 4P2

k 100P4 l 200P3

A 11605 B 10254

C 14 250 D 12 540

E 13 692

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Combinatorics

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b Determine the number of permutations if theorder of the comic books in each bundle doesnot change.

a In how many ways can thisbe done if there are norestrictions on where indi-vidual items are to be placed?

10 A novelty sports day carnivalinvolves 10 competitors. Aprize is given to the winner ofthe first race who then cannottake part in the remainder ofthe races. The winner andrunner-up of the second raceare awarded prizes and arethen eliminated from theremainder of the events.

Similarly, the first threeplace-getters of the third raceare given prizes and must dropout of the competition. This iscontinued until the number ofcompetitors remaining is the same as the number of prizes to be awarded. How manydifferent ways can prizes be awarded?

11 There are three separate bundles of reading material comprising 4 comics, 2 novelsand 3 magazines. They are placed together to form one bundle.


Permutations involving restrictionsIdentical objectsSo far our study of permutations has been based on the assumption that the objectsarranged were all different (distinguishable). We will now examine the situation whensome of the objects are identical (indistinguishable).

A scrabble player has the following letter tiles: A, A, A, B, C, D, E. If the As weredistinguishable, we might consider them to be A1, A2, A3 and could begin to list thepossible arrangements of the 7 letters as follows:

A1A2A3BCDE ....... ....... ....... and so onA1A3A2BCDE ....... ....... ....... and so onA2A1A3BCDE ....... ....... ....... and so onA2A3A1BCDE ....... ....... ....... and so onA3A1A2BCDE ....... ....... ....... and so onA3A2A1BCDE ....... ....... ....... and so on

Without listing them all, we can calculate there are 7P7 = 7! = 5040 possible arrange-ments. But the As are not distinguishable. So, really, the arrangements listed above areall the same as AAABCDE, which counts as one arrangement.

Because there are 3 As we have 3! = 6 times too many arrangements, hence we needto divide 5040 by 6.

This means there are only = = = 840 different arrangements or permu-

tations of 7 objects where 3 of them are identical.

In general, the number of arrangements of n objects, n1 of which are identical,

is given by

Extending this formula we have:

The number of ways of arranging n objects, which include n1 identical objects of one type, n2 identical objects of another type, n3 identical objects of yet another type and so on is:

P7

7

3!------- 7!

3!----- 5040

6------------

n!n1!--------

n!n1!n2!n3!…-------------------------------

In how many ways can 4 identical red marbles and 3 identical blue marbles be placed in a row?

THINK WRITE

There are 4 + 3 = 7 objects altogether.The number of ways the blue marbles can be arranged is 3!, and the red marbles 4!

Substitute the values in the formula.

Number of ways

=

= 35

1

2

37!

4! 3!×----------------

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Grouped objectsIn how many ways can the letters A, B, C, D be positioned in a row? We knowthat this can be done in 4! ways, but what would be the answer if the question hadbeen: ‘In how many ways can the letters A, B, C, D be positioned in a row if Aand B must be next to each other?’ The number of arrangements will clearly beless than 4! because of the restriction imposed on A, B. The figure below showsthe 4! possible arrangements of A, B, C, D which include the 12 ways A and B aretogether.

If A and B are to be together we consider the problem to be one of arranging 3objects, say X, C and D, where one of the objects, X is the group containing A and B.

The figure below shows that there are 6 arrangements with A and B together.

The 3 objects can be arranged in 3! ways and within the group A and B can them-selves be arranged in 2! ways (namely AB and BA). The multiplication principle is nowused so that the number of arrangements when A and B are together is 3! × 2! = 12.

Now consider the permutations if A, B, C must be together. Again, we view theletters as consisting of two objects X and D, where X is the group of letters A, B and C.Thus we have two objects to arrange in 2! ways as shown below.

Among themselves the letters A, B, C contained in X have 3! different arrangementsas shown below.

Therefore the total number of arrangements when A, B and C are together is2! × 3! = 12.

We can generalise this approach to include any number of groups of objects that arerequired to be together.

If n objects are to be divided into m groups with each group havingG1, G2, G3, . . . Gm objects respectively, the number of arrangements is given bym! × G1! × G2! × G3! × . . . × Gm!

A B C D B A C D C B A D D B C AA B D C B A D C C B D A D B A CA C B D B C A D C A B D D C B AA C D B B C D A C A D B D C A BA D B C B D A C C D B A D A B CA D C B B D C A C D A B D A C B

A B C D C A B D C D A BA B D C D A B C D C A B

X D D X

A B C D D A B CA C B D D A C BB A C D D B A CB C A D D B C AC A B D D C A BC B A D D C B A


The letters of the word TABLES are placed in a row. How many arrangements are possible if the letters T, A and B must be together and the letters E and S must be together?

THINK WRITE

Consider the letters T, A and B as one object (group). There are 4 objects to be arranged, namely the TAB group and the letters L, E and S.The letters in the group TAB can be arranged in 3! ways.Use the multiplication principle. Number of arrangements = 4! × 3!

= 144

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2

3

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Five cars — a Toyota, a Ford, a Commodore, a Corolla and a BMW — are to be parked side by side. In how many ways can this be done if the Toyota and BMW are not to be parked next to each other?THINK WRITE

The five cars can be arranged in 5! ways without restriction.

Number of ways of arranging 5 cars = 5!

Calculate the number of arrangements where the Toyota and BMW are together (4! × 2!).

Number of ways where the Toyota and BMW are not together

= 5! − 4! × 2!Subtract from the unrestricted number of arrangements the number of ways the two cars are together.

= 120 − 48= 72

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2

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The letters of the word REPLETE are arranged in a row. In how many ways can this be done if the letters R and P must not be together?THINK WRITE

Find the number of unrestricted arrangements of the 7 letters and consider that there are 3 identical Es.

Number of ways of arranging 7 letters with 3 Es

=

Calculate the (restricted) number of ways R and P are together. Consider R and P as one object so there are 6 objects to arrange. There are three Es to consider (3! ways). R and P can be arranged in 2! ways within their group.

Number of ways where R and P are not together

= −

Subtract the number of ways with R and P together from the total number of arrangements.

= 840 − 240= 600

1

7!3!-----

2

7!3!----- 6! 2!×

3!-----------------

3

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Permutations involving restrictions

1 In how many ways can 5 identical white beads and 4 identical yellow beads bearranged in a straight line?

2 Three 5 cent coins, two 10 cent coins and six 20 cent coins are to be placed side byside. Determine how many ways this can be done.

3The number of permutations using the letters of the word LOOPHOLE is:

4 The toy set shown in the photo consists of a number of indistinguishable brownhorses, 1 white horse, a cowboy and 3 indistinguishable black horses. In how manyways can they be placed end to end?

A 3520 B 3360 C 4000 D 4150 E 3840

remember1. An ordered collection of objects is a permutation or an arrangement.2. The number of ways of arranging n objects which include n1 identical objects

of one type, n2 identical objects of another type, n3 identical objects of yet another type, and so on is:

3. n objects divided into m groups, with each group having G1, G2, G3, . . . Gm objects respectively, has m! × G1! × G2! × G3! × . . . × Gm! arrangements.

n!n1!n2!n3! . . . ---------------------------------

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C h a p t e r 1 1 C o m b i n a t o r i c s 4975 How many different 6-digit numbers can be obtained using the digits 4, 6, 7, 6, 6, 4?

6 A party-light kit consists of 20 coloured globes connected to each other in a straight line.a If there are 5 red globes, 6 blue globes, 7 yellow

globes and a number of green globes as shown at right, find how many different arrangements of coloured globes are possible.

b How many different permutations of coloured globes are there if the first and last globes must both be red?

7 Find how many arrangements are possible altogether when the letters of the wordCHAIR are placed in a row and C and H are to be next to each other.

8 The digits 5, 3, 6, 2 and 7 are used to make a 5-digit number. How many differentnumbers are possible if the digits 3, 2 and 7 must be together?

9 Maria, Steven, James, Sofia, Nin and Alfredo are standing next to each other. Calculatehow many ways this can be done if Maria and James are not to stand next to each other.

10 Establish the number of ways in which 7 different books can be placed on a bookshelfif 2 particular books must occupy the end positions and 3 of the remaining books arenot to be placed together.

11Ten athletes line up for a race. The number of per-mutations when three of the athletes — Sam, Troyand Pablo — would be next to each other is:A 3 628 800B 1 209 600C 241 920D 5 443 200E 4 838 400

12 A carpenter has 3 identical hammers, 5 different screwdrivers, 2 identical mallets,2 different saws and a tape-measure. She wishes to hang the tools in a row on a toolrack on the wall. In how many ways can this be done if the first and last positions onthe rack are to be mallets and the hammers are not to be next to each other?

13 Decide in how many ways the letters of the word ABRACADABRA can be arrangedin a row if C, R and D are not to be together.

14The number of ways the letters of the alphabet can be placed in a straight line with therestriction that the letters of the sentence UP THE BIG SKY WORLD must not betogether is:A 26! − 11!16! B 26! + 16! C 16! − 8!D 6!16! E 6!16!26!

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Arrangements in a circleAnna, Betty and Lin stand on the circumference of a circle painted on the school’splayground. How many different arrangements are there?

The figure below shows the two arrangements for the girls’ positions on the circle.

Notice that Anna is ‘locked’ in position to provide a reference point and Betty andLin are arranged around Anna in 2! (= 2) ways.

Compare this with the 3! (= 6) arrangements in a line.

Susie now joins the group to make 4 people in a circle.We can designate any of the 4 girls in the circle as our ‘start’ by ‘fixing’ one person

(in this case, Anna) in one position and arranging the remaining girls around her. Thisreduces by one person the number of girls to arrange.

There are 3! (= 6) ways of arranging 4 people in a circle. Compare this with 4! (= 24)arrangements in a line.

In general:

n distinguishable objects can be arranged in a circle in (n − 1)! ways.

Betty

Lin Lin Betty

AnnaAnna

ABL BAL BLA LBA ALB LAB

(A is Anna, B is Betty, L is Lin)

(A is Anna, B is Betty, L is Lin, S is Susie)

A

B L

S

A

B S

L

A

L B

S

A

L S

B

A

B L

S

A

S B

L

In how many ways can these five children be arrangedin a circle?


In how many ways can the vowels of the alphabet be arranged in a circle?

THINK WRITE

The vowels are a, e, i, o, u. Therefore, there are 5 objects to arrange.

n = 5

Use (n − 1)! with n = 5. Number of ways = (n − 1)! Number of ways = (5 − 1)! Number of ways = 4! Number of ways = 24

1

2

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Calculate the number of arrangements that are possible using the letters of the word UNUSUALLY in a circle.

THINK WRITE

There are 9 letters, so use n = 9 with(n − 1)!We need to consider repetition of letters. There are three Us and two Ls.

Use the formula for repetition of objects.

Number of arrangements =

Number of arrangements = 3360

1

2

38!

3! 2!×----------------

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In how many ways can 6 people sit around a table if two particular people must be seated next to each other?

THINK WRITE

Consider the two people required to sit together as being one object. So there are 5 objects to arrange in a circle.The two people can be arranged in 2! ways.Use the multiplication principle. Number of ways = 4! × 2!

Number of ways = 48

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2

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remember1. n distinguishable objects can be arranged in a circle in (n − 1)! ways.2. The same methods are applicable to arrangements in a circle as the methods

used for indistinguishable objects when there are restrictions on the possible arrangements.

remember


Arrangements in a circle

1 Calculate the number of ways in which the letters of the word PENCIL can bearranged in a circle.

2

Eight children hold hands to form a circle in the playground. The number of ways thiscan be done is:

3 Determine the number of arrangements in a circle that are possible when the letters ofthe word EXCELLENT are used.

4 A child uses coloured dots on paper to represent the hour marks of a clock face. Howmany permutations are possible if there are 4 orange dots, 5 white dots, 2 black dotsand 1 purple dot?

5 A family of 3 adults, 2 boys and 3 girls are sitting around a circular dinner table. Findthe number of seating positions that are possible if the 3 boys are to be together.

6 A special pizza consists of 10 slices with different toppings used. If 2 slices areCapricciosa, 5 slices are Supreme and 3 slices are Ham and Pineapple, how many dif-ferent arrangements of pizza slices are possible?

7 A manufacturer of merry-go-rounds uses 8 identical wooden horses, 4 identicalplastic motorbikes and 2 different miniature cars. They are all equally connectedaround the rim of a circular moving base. Establish how many different arrangementsthere can be if the 2 cars are not to be placed in consecutive positions.

8

Ten owners of pedigree dogs will enter the arena to parade their dogs by walkingaround a circular track. Unfortunately, 3 particular dogs cannot get along together andso cannot parade if all 3 are next to each other. There appears to be no problem if anytwo of this group of 3 dogs are together. The number of ways of avoiding thisproblem is:

9 In how many ways can the letters of the word POTATOES be arranged in a circle?

10

The letters of the word FULFILLED are to be arranged in a circle. The number ofarrangements possible when U and E are together or when U, E and D are together is:

11 To publicise the venue, a hotel manager gave a gift to each of 12 prominent business-people as they went into the conference room and seated themselves at a round tableto begin discussions. The gifts comprised 4 fountain pens, 5 pocket electronic organ-isers and 3 calculators. Calculate what fraction of the possible unrestricted arrange-ments is the number of arrangements that has 4 businesspeople who have been givena fountain pen sitting next to each other.

A 6280 B 5400 C 3680 D 4320 E 5040

A 3 588 480 B 3 870 720 C 3 628 800 D 3 386 880 E 3 540 650

A 3140 B 1940 C 2000 D 1200 E 1850

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Combinations using nCrTaking combinations involves the selection of r objects from n objects without con-sideration for the order of the elements. For example, the number of permutations oftwo letters selected from the letters A, B, C, D is 4P2 = 12. The arrangements are:

AB AC AD BC BD CDBA CA DA CB DB DC

If we are not concerned with order, there are only 6 selections:

AB AC AD BC BD CDThe 2! ways of arranging the elements of the 2-element subgroup is not considered.

Now consider the selection of 3 letters from A, B, C, D. The number of ordered sub-sets is 4P3 and each subset of 3 elements can be arranged in 3! ways. Therefore 4P3 isthe number of unordered subsets of 3 objects multiplied by the number of ways the 3objects can be arranged.

In general terms it can be stated that nPr is the number (C) of unordered groups of robjects multiplied by the number of arrangements (r!) of r objects.

That is, nPr = C × r! so that .

Now by the definition of nPr = we have:

The number of combinations is usually denoted by nCr or so we have:

1. The number of combinations of r objects selected from n objects is:

where n, r are natural numbers and r ≤ n.

2. or

The function nCr is a standard mathematical function to be found on scientific andgraphics calculators.

To calculate nCr using the TI–83 on the home screen, type in the value of n first, then do the following:1. Press select PRB and 3:nCr.2. Type the value of r, and press .

The screen dump at right is for finding 10C3.

CnPr

r!--------=

n!n r–( )!

------------------

Cn!

n r–( )!------------------ r!÷ n!

r! n r–( )!-----------------------= =

nr

nr

nCrn!

r! n r–( )!-----------------------= =

nPrnCr r!×= n

r nCr

nPr

r!---------= =

Graphics CalculatorGraphics Calculator tip!tip! Combinations

MATHENTER


Special cases1. If r = 0, then nCr = nC0 = .

This implies that there is one way of selecting 0 objects from n objects.

2. If r = n, then nCr = nCn = .

There is one combination of n objects taken from n objects.

3. If r = 1, then nCr = nC1 = .

If objects are taken one at a time from n objects, there are n combinations.From 1. and 2. we conclude that nC0 = nCn.

This is an instance of the general case that:

For example, 7C4 =

and 7C3 =

so 7C4 = 7C3

n!0! n 0–( )!------------------------ n!

1 n!×-------------- 1= =

n!0! n n–( )!------------------------ n!

n 0!×-------------- 1= =

n!1! n 1–( )!------------------------ n! n 1–( )!

1 n 1–( )!×---------------------------- n= =

nCrnCn r– or n

r n

n r– = =

7!4!3!----------

7!3!4!----------

Evaluate 6C2.

THINK WRITE

Use the definition 6C2 =

6C2 =

A calculator which has the function nCr , where n = 6 and r = 2 may be used.

6C2 = 15

1 nCrn!

r! n r–( )!-----------------------= 6!

2! 6 2–( )!------------------------

6!2! 4!×----------------

2

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Evaluate .

THINK WRITE

Express in factorial form.

100! = 100 × 99 × 98!

Evaluate. =

= 4950

100

98

1100

98 100

98 100!

98!2!-------------=

100 99 98!××98! 2 1××

-----------------------------------=

2

3100 99×

2 1×---------------------

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In how many ways can a committee of 2 boys and 3 girls be formed from a group consisting of 5 boys and 8 girls?THINK WRITE

Select 2 boys from 5 boys.Select 3 girls from 8 girls.Use the multiplication principle. Number of ways = 5C2 × 8C3

Number of ways = 10 × 56Number of ways = 560

123

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A committe of 6 is to be formed from a group of 5 men and 4 women.a How many committees can be formed?b How many committees contain 3 men and 3 women?c How many committees contain at least 4 men?

THINK WRITE

a Use n = 9 and r = 6 with nCr. a Number of committees = 9C6 Number of committees = 84

b Select 3 men from 5 men and 3 women from 4 women.

b Number of committees= 5C3 × 4C3 = 10 × 4= 40

Use the multiplication principle.(‘and’ situation)

c At least 4 men means 4 men and 2 women or 5 men and 1 woman.Select 4 men and 2 women.Select 5 men and 1 woman.Sum the answers because the 2 events are mutually exclusive.(‘or’ situation)

c Number of committees= 5C4 × 4C2 + 5C5 × 4C1 = 5 × 6 + 1 × 4= 30 + 4= 34

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234

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remember1. A combination is the arrangement of objects where order is not important.2. The number of combinations when r objects are selected from n objects is

denoted by nCr or .

3. nCr = = nCr = = nCr = nCn − r

nr

nr

n!r! n r–( )!----------------------- n

r

nPr

r!--------

remember


Combinations using nCr

1 Calculate:

2 Evaluate each of the following.

3 Determine the value of each of the following.

4

The value of 2 × 4C2 + 3 × 5C3 is:

5 Calculatea 3C1 and 3C2 b 4C1 and 4C3 c 5C2 and 5C3 d 9C4 and 9C5

6 Copy and complete:a 20C7 = 20C__ b 100C9 = 100C__

7 In how many ways can 5 objects be chosen from 12?

8 How many combinations are possible if 2 numbers are chosen from 6 in a mini-lotto game?

9 A student must choose 5 types of party food from the following list: sausage rolls,potato crisps, fairy bread, party pies, cheezels, cocktail frankfurts and celery sticks.How many different combinations may be chosen?

10 A committee of 6 must be chosen from a meeting of 30 people. How many differentcommittees are possible?

11 In how many ways can a group of 3 boys and 4 girls be formed from a groupconsisting of 4 boys and 6 girls?

12 A magazine pile in a waiting room contains 6 glamour magazines and 7 computermagazines. In how many ways can a patient choose 2 glamour and 3 computermagazines to flick through during a lengthy wait?

a 5C2 b 4C3 c 6C1d 8C0 e 9C9

a b c

d e

a b c

d e

A 42 B 90 C 80 D 94 E 70

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6461

3834

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Combinatorics

C h a p t e r 1 1 C o m b i n a t o r i c s 50513 A school offers 10 sience subjects and 15 humanities subjects to prospective Year 12

students. In how many ways may a student choose 4 science and 2 humanitiessubjects?

14 How many 10-card hands containing exactly 7 hearts and 3 spades are possible froma standard 52-card deck?

15 A committee of 5 parents is to be established from a group of 6 men and 4 women.a Find how many different committees can be formed.b How many different committees are possible consisting of 3 men and 2 women?

16 A retirement home organises an adventure camp for its residents, who must choose2 or 3 activities from the following: paragliding, abseiling, skydiving and bungeejumping. In how many ways may a group of activities be chosen?

17 An ice-cream vendor offers chocolate, strawberry and vanilla ice-creams with one,two or three scoops. How many different ice-creams are possible?

18 A mixed basketball squad of 10 must be chosen from a group of 8 women and 6 men.How many squads are possible:a without restriction?b if the squad contains 6 women and 4 men?c if the squad must contain at least 6 women?d if the squad contains all the men?

19 A sub-committee of 3 people must be chosen from a group of 9 teachers (whichincludes the principal). How many sub-committees may be chosen:a that contain the principal?b that do not contain the principal?

20 To win LottoMania the 5 numbers entered on the player’s entry ticket must be thesame as 5 numbers that are randomly selected from the numbers 1 to 30.a How many different entries are possible?b What is the percentage increase in the number of possible combinations if the

numbers are randomly selected from the numbers 1 to 35?

21

A painter has 7 colours at her disposal. The number of additional colours that can beobtained by mixing equal amounts of any number of the 7 colours is:

22 Determine the number of ways in which 8 people can be divided into 2 equal groups.

23

The number of ways in which 10 objects can be divided into 2 unequal groups is:

A 100 B 128 C 5040D 5120 E 120

A 385 B 835 C 950D 640 E 565

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Pascal’s triangleThe French mathematician, physicist and philosopher Blaise Pascal (1623–62) proposed a useful way of presenting results derived from his extensive work in the field of probability. It has become known as Pascal’s triangle.

Pascal’s triangle, from row 0 to row 5 appears below.

1 Copy and extend Pascal’s triangle up to row 10, being careful to allow space for later (wider) rows.

2 Copy and complete the following table. Note that term numbering starts at zero in each row.

3 How could you use Pascal’s triangle to find the value of a 50C13? b nCr?4 Find the sum of numbers in each row up to row 10. Suggest a formula for

calculating the sum of numbers in the nth row of Pascal’s triangle.5 Pascal’s triangle contains many other patterns. Can you find any?6 Expand:

a (x + y)2

b (x + y)3

c (x + y)4.What connection is there with your answers and Pascal’s triangle?

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Pascal’striangle

Note the use of row 0 as reference for the first row.

1

2

6

1 1

3 3

10 10

11

44

1

1 1

11

1

5 5

Row

0

1

2

3

4

5

Term 0 Term 1 Term 2

Number inPascal’striangle

Row(n)

Term number(r)

(Count 1 as the 0th termin each row)

nCr

or

106

56120

210

5

79

3

62

n

r


Applications to probabilityWe define the probability of an event to be:

The methods we have used to calculate permutations and combinations can also beapplied to problems involving probability.

Pr(event)number of favourable outcomes

total number of possible outcomes----------------------------------------------------------------------------------=

Romina makes a guess as to which 2 of 10 swimmers will come first and second in a race. What is the probability that her guess will be right?

THINK WRITE

Calculate in how many ways 2 swimmers can be chosen from 10 swimmers, where the order is taken into account. Use nPr where n = 10 and r = 2.

The number of favourable outcomes is 1 because Romina makes only one guess.

Pr(correct guess) =

Pr(correct guess) =

Use the formula for probability.

1

21

10P2-----------

145------

3

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A computer randomly interchanged the letters of the word CREATIONS. Find the probability that the letters A and T end up together.

THINK WRITE

If A and T are together, treat them as one object, therefore we have 8 objects.

AT can be arranged in 2! ways.

Find the number of ways the 9 letters can be arranged.

Pr(A and T are together) =

Use the formula for probability. Pr(A and T are together) =

1

2

38! 2!×

9!----------------

429---

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A committee of 5 people is to be formed by choosing members from a group of 6 men and 4 women. What is the probability that the committee will consist of 3 men and 2 women?THINK WRITE

Calculate the number of ways in which 3 men can be selected from 6 men and 2 women can be chosen from 4 women.Use the multiplication principle to establish the number of favourable outcomes of the committees consisting of 6 men and 4 women.Determine the number of ways in which 5 people can be selected from the group of 10 people.Use the formula for probability.

Pr(3 men and 2 women)

=

=

=

1

2

3

4

6C34C2×

10C5-------------------------

20 6×252

---------------

1021------

28WORKEDExample

Eight people randomly seat themselves about a circular table. What is the probability that 3 particular people will be sitting next to each other?

THINK WRITE

Treat the 3 people as one object, therefore there are 6 objects to arrange.Use the formula (n − 1)! for arrangements in a circle for the situation where the 3 people are together.Calculate the total number of possible outcomes for the 8 people, using (n − 1)!Use the formula for probability.

Pr(3 particular people seated together)

=

=

=

= 0.143

1

2

3

46 1–( )! 3!×

8 1–( )!------------------------------

5! 3!×7!

-----------------

7205040------------

29WORKEDExample

rememberThe probability of an event is given by:



remember


Applications to probability

1 Jenny, Hakan and Miriam are competing in a car race against 5 other drivers. Theirfriend Mary predicts that they will cross the finish line first, second and third respec-tively. What is the probability that Mary is right?

2 The letters of the word PRODUCE are randomly reordered. Calculate the probabilitythat the letters P and E will be together.

3 Six people selected from 5 men and 7 women are to form a committee. Work out theprobability that the committee will consist of 3 men and 3 women.

4

The letters A, B, C, D, E and F are randomly placed in a row. The probability that theletters A and B will occupy the first and second positions respectively is:

5 Six cards are randomly distributed from a standard pack of 52 playing cards. Deter-mine the probability that exactly one of the 6 cards is a queen.

6 From a toy-set consisting of 4 dolls and 5 clowns, 2 toys are chosen at random. Findthe probability that the 2 toys are 2 clowns or 2 dolls.

7From a group of 3 children and 8 adults, 5 will be chosen to receive prizes. Theprobability that 2 children and 3 adults will be awarded a prize is:

8 A group comprising 6 people is sittingaround a table. Find the probabilitythat two particular people are sittingnext to each other.

A B C D E

A B C D E

11HWORKEDExample

26

EXCEL Spreadsheet

Combinatorics

Mathcad

CombinatoricsWORKEDExample

27

WORKEDExample

28


115------ 1

3--- 1

30------ 1

6--- 2

3---


33

85

115

-----------------

31

81

113

118

-----------------------

32

83

+

115

-----------------------

32

83

115

-----------------

32

83

113

118

+-----------------------------

WORKEDExample

29


9 Ten people are seated at a circular dining table. Find the probability that twoparticular people will be sitting next to each other.

10Six mothers and their 6 daughters randomly arrange themselves in a circle. Theprobability that Susan is next to her daughter Jeanette is:

11 Four letters are randomly selected from the word ENCYCLOPAEDIA. Find theprobability that one letter E will occur in the selection of 4 letters.

12 A school captain and 2 vice-captains are to be chosen from a group of 5 boys and 6 girls. What is the probability that all 3 positions will be taken by:a boys? b girls?c two boys and one girl? d at least two girls?

13 Four colours are randomly picked from the 7 different colours of the rainbow. Calcu-late the probability that yellow will not be one of the colours chosen.

14 To win Division 1 in the game of Tattslotto, the player must have the same 6 numbers(in any order) as those that are randomly drawn from the numbers 1 to 45. A Division2 prize requires that the player’s ticket must have 5 of the 6 winning numbers and aDivision 3 prize requires that the player has 4 of the 6 numbers drawn. Calculate theprobability that the player’s 6 numbers will contain at least a Division 3 prize.

15 Five letters are randomly selected from the letters of the word HOLIDAYS and placedin a row. Calculate the probability that the first letter chosen is a consonant.

16

Inside a box are n objects of which m are white. If r objects are randomly taken out ofthe box and placed in a row, the probability that the first object is white is:

17

A 5-digit number is randomly formed using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. If adigit cannot be used more than once in the number,a the probability that the number is even is:

b the probability that the number is between 30 000 and 50 000 is:

18 A debating team of 6 people is to be formed from a group consisting of 5 males and6 females.a What is the probability that the team will consist of at least one male?b What is the probability that the team will have at least four females?

A B C D E

A B C D E

A B C D E

A B C D E


14--- 5

6--- 2

3--- 1

12------ 2

11------


nm---- m

n!----- m

n---- m n+

n!------------- n m–

n-------------


9100--------- 57

195--------- 4

9--- 35

78------ 14

63------

2057------ 19

73------ 2

9--- 35

78------ 14

63------


Addition principle• The addition principle states that if two operations can be performed in A or B ways

respectively, then both operations can be performed together in A + B ways.

Multiplication principle• The multiplication principle states that if two operations can be performed in A and

B ways, then both operations can be performed in succession in A × B ways.

Factorials• The factorial of a positive whole number n is defined as:

n! = n × (n − 1) × (n − 2) × (n − 3) × . . . × 3 × 2 × 1 with 0! = 1

Permutations• A permutation is the arrangement of objects in a definite order. The multiplication

principle is commonly used in calculating the number of possible permutations.

• nPr =

• The number of ways of arranging n objects which include n1 identical objects of one type, n2 identical objects of another type, n3 identical objects of yet another type and so on is:

• n objects divided into m groups, with each group having G1, G2, G3, . . . Gm objects respectively, has m! × G1! × G2! × G3! × . . . × Gm! arrangements.

Arrangements in a circle• n distinguishable objects can be arranged in a circle in (n − 1)!• The same methods are applicable to arrangements in a circle as the methods used

for indistinguishable objects when there are restrictions on the possible arrangements.

Combinations• The number of combinations when r objects are selected from n objects is denoted

by nCr or . nCr = = or nCr = nCn − r

Applications to probability• The probability of an event:

Pascal’s triangle

• nCr = = the rth term in row n

of Pascal’s triangle(Start numbering of terms and rows at zero.)

Term 0 Term 1 Term 2

summary

n!n r–( )!

------------------

n!n1!n2!n3!…------------------------------

nr

nr

n!r! n r–( )!-----------------------

nPr

r!--------



1

2

6

1 1

3 3

10 10

11

44

1

1 1

11

1

5 5

Row

0

1

2

3

4

5

n

r


Multiple choice

1 Samantha can get to work by walking, by taking her car or by using public transport (train, tram, bus, taxi). The number of different ways she can get to her work is:

2 Malcolm is guessing someone’s house number. He knows that the number is an odd number and is between 30 and 60. Assuming that the same guess is not made twice, the maximum number of guesses he can make is:

3 The total number of 2-digit, 3-digit and 4-digit odd numbers that can be formed using the digits 6, 4, 5, 2, 1 when a digit cannot be used more than once is:

4 The value of is:

5 The value of 9! − 7! is equivalent to:

6 The value of 7P5 is:

7 The number of permutations using the letters of the word MISSISSIPPI is:

8 Five letters are chosen from the letters of the word WATERING and placed in a row. The number of ways in which this can be done if the last letter is to be W is:

9 A family, consisting of a mother, father, 3 sons and 4 daughters, lines up for a photograph. How many ways can this be done if the daughters must be together?

10 Eleven members of a cricket team are to be seated in a circle. The number of possible arrangements is:

A 5! B 10! C 11! D E

11 The letters of the word MUSICAL are to be arranged in a circle. If the letters U and S must not be together, the number of possible arrangements is:

A 3 B 5 C 4 D 6 E 2

A 15 B 20 C 30 D 45 E 25

A 200 B 80 C 170 D 120 E 128

A undefined B 1000 × 999 C 996! D 998 × 997 E 998 × 997 × 996

A 71 × 7! B 2! C 7! × 9 D 8! E 7 × 8!

A 21 B 42 C 2520 D 1008 E 5040

A 4! B 11! C D 1!2!4!4! E

A 840 B 2520 C 210 D 40 E 625

A 9! B 6!4! C 5!4! D 2!3!4! E 10!

A 480 B 718 C 1440 D 3600 E 5038

CHAPTERreview

11A

11A

11B

11C998!996!-----------

11C

11E 11!4!4!2!---------------- 1

4!-----

11E

11E

11!10!-------- 10!

11--------

11F

C h a p t e r 1 1 C o m b i n a t o r i c s 51312 Joanna has decided to study part time at university. Her course requires that she undertake at

least 2 subjects for the year. If 4 subjects are being offered, the number of subject combinations is:

13 Four pieces of fruit are selected from a box containing 5 oranges and 6 apples. The number of selections that contain at least 2 oranges and 1 apple is:

14 Five letters are randomly selected from the word ENERGISE. The probability that the letter E will appear in the group of 5 letters is:

Short answer1 There are 7 airlines that have flights from Australia to Singapore, 6 airlines that offer flights

from Singapore to Europe and 5 airline companies that service the route from Europe to America. Determine the number of different travel arrangements possible to get from Australia to America via Singapore and Europe.

2 Find the number of ways at least one letter can be selected from the letters of the word CLOCK.

3 Seven people form a queue to board a bus. How many different queues are possible?

4 The digits 3, 5, 6 and 8 are used to form numbers greater than 100. If a digit may be used once only and not all digits have to be used, how many different numbers can be formed?

5 Seven different books are to be placed on a shelf. If a particular book must occupy the first position, find the number of permutations possible.

6 A number containing at least 3 digits is to be formed using digits taken from 8, 4, 3, 6 and 7. If a digit may be used more than once, how many different numbers can be made?

7 The 4 fastest runners in a race will qualify for the finals. If there are 11 competitors, determine the number of different ways in which the race can finish.

8 Evaluate 9! + 8! − 6! + 3 × 2!

9 What is the value of 5 × 4P2 − 3 × 6P1?

10 In how many ways can first, second and third prizes be awarded to 12 people competing in a marathon?

11 In how many ways can 6 identical spoons, 5 identical forks and 4 identical knives be arranged in a row?

12 Find the number of ways the letters of the word ARRANGEMENT can be placed in a row.

13 Anna, Belinda, Chien, Deanna and Erica are lining up for concert tickets. If Belinda and Deanna do not want to be next to each other, what is the number of possible queues?

14 Ten children are arranging themselves in a circle. Calculate the number of ways this can be done if three particular children are not to be next to each other.

15 Two students from a group of 8 students are to be class captain and vice-captain. From the remaining candidates, two will become class monitors. Find the number of ways this can be done.

A 36 B 24 C 15 D 11 E 20

A 210 B 150 C 60 D 90 E 110

A B C D E

11G

11G

11H58--- 15

56------ 3

8--- 1

8--- 55

56------

11A

11A11A11B

11B

11B

11B

11C11D11D

11E

11E11E

11F

11G


16 A class consists of 24 students. If an initial group of 4 must be chosen to go for a measles injection, how many different combinations may be selected for that group?

17 A team of at least 2 people must be chosen from a group of 5 mountaineers to mount a rescue mission. How many different teams may be chosen?

18 A committee of 5 people is to be established using members from a group of 6 men and 7 women. What is the probability that the committee will contain 2 men and 3 women?

19 The letters of the word FEATURING are randomly rearranged. Find the probability that the letters of the word FEAT are together, though not necessarily in the order shown.

Analysis1 From a group of 20 female students, 2 female staff, 18 male students and 3 male staff, a

committee of 6 is to be formed. a Find the number of different committees if:

i There are no restrictions.ii All committee members must be students.iii One female and one male staff member must be on the committee.iv There is an equal number of males and females on the committee.v One particular student must be on the committee.vi One particular student must not be on the committee.vii The committee must comprise 2 male staff members, 2 male students, 1 female staff

member and 1 female student.b Find the probability that:

i only students are selected for the committeeii all the staff are selected for the committeeiii exactly 2 staff and 4 students are selected.

2 Two women and three men approach an ATM at the same time.a How many different queues are possible if the position of each person in the queue is

taken into account?b How many queues of at least two people are possible if the position of each person in the

queue is not taken into account?

11H

11H

testtest

CHAPTERyyourselfourself

testyyourselfourself

11

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