19 19.2permutation 19.3combination chapter summary case study permutation and combination...

19

19.2 Permutation

19.3 Combination

Chapter Summary

Case Study

Permutation and Combination

19.1 Counting Principle

P. 2

Andy joins a summer camp.On the first day, each team member has to shake hands with all other members in the same team once.

Case StudyCase Study

If there are 4 members in each team, what is the total number of times each team must shake hands?

You may try to work it out by counting but it won’t be easy.

If each team member has to shake hands with all other members once, then how many times does each team need to shake hands?

Let the 4 team members be A, B, C and D respectively. From the figure, we then have the following possibilities:

AB AC AD BC BD CDTherefore, the total number of times each team must shake hands is 6.

P. 3

Consider the following situation. Mr. Lee plans to go to Macau this weekend. He can choose the ferry service offered by company A or company B. The number of ferry trips offered by company A and company B are 32 and 28 respectively.

19.1 19.1 Counting PrincipleCounting Principle

A. A. Addition Rule in the Counting PrincipleAddition Rule in the Counting Principle

Therefore the total number of ferry trips that he can choose from is

32 28 60

In this section, we will learn the concept of the counting principle and how to apply two counting rules to determine the number of possibilities that exist in a given situation.

ferry tips from company A ferry tips from company B

P. 4



We find the answers by applying the addition rule in the counting principle which is stated as follows:

Addition Rule in the Counting Principle If there are k different choices to finish a task and there are n1 ways to finish the 1st choice, n2 ways to finish the 2nd choice, ... , and nk ways to finish the kth choice, then the total number of ways to finish the task n1 + n2 + … + nk.

P. 5

Example 19.1T

Solution:


The opening ceremony of a library will be held on a day either in October or November. However, the ceremony will not be held on 1st October. How many choices do they have for the day of the ceremony?


If they choose a day in October, then they have 30 choices. If they choose a day in November, then they have 30 choices.

The number of choices they have3030

60

P. 6


BB. . Multiplication Rule in the Counting PrincipleMultiplication Rule in the Counting Principle

Suppose there is only one task. If the procedure for finishing thetask involves several steps, we have to use the multiplication rule instead of the addition rule to count the number of ways to finish the task.

A fast food shop offers hamburger sets which consist of a hamburger and a drink. A customer may choose one filling for the hamburger and a drink from the menu.

We can list all the possible ways of choosing a hamburger set as below.

Beef + Coffee Beef + Tea Beef + Orange juice Cheese + Coffee Cheese + Tea Cheese + Orange juice Chicken + Coffee Chicken + Tea Chicken + Orange juice Sausage + Coffee Sausage + Tea Sausage + Orange juice

Therefore, the number of ways of choosing a hamburger set is4 3 12

number of ways of choosing a filling from the 4 choices

number of ways of choosing a drink from the 3 choices

P. 7



In general, if we want to perform a task with more than one step and each step will not affect the others, we can apply the multiplication rule in the counting principle.

Multiplication Rule in the Counting Principle Suppose a task can be divided into k steps and each step will not affect the others. If there are n1 ways to finish the 1st step, n2 ways to finish the 2nd step, ... , and nk ways to finish the kth step, then the total number of ways to finish the task n1 n2 … nk.

P. 8



Consider the example of choosing a book from the shelf.

If, instead of choosing only one book, Lily has to choose a book from each of the three categories, then how many choices are there?

We can divide the task into three steps:

Step 1: Choose a literary book from 10 choicesStep 2: Choose a fiction book from 15 choicesStep 3: Choose a science book from 12 choices

k 3

n1 10

n2 15

n3 12

Therefore, the number of choices is 10 15 12 1800.

P. 9

Example 19.2T


A basketball tournament lasts for 4 days. On each day, there are 8 matches. How many matches are there in total? Solution:


By the multiplication rule, the number of matches

8432 The first step is to choose a

day and the second step is to choose a match.

P. 10

Example 19.3T


The order numbers from a pizza shop are made up of a non-vowel letter and a three-digit number from 101 to 999. How many possible order numbers are there?

Solution:


By the multiplication rule,

the number of order numbers )100999()526(

879 18

P. 11

Example 19.4T

Solution:

5 coins are put into 4 boxes. If each box can contain at most 5 coins, in how many ways can the coins be put into the boxes?

We can interpret the ‘step’ as ‘choosing a coin’.

the number of ways of putting the coins into the 4 boxes 44444

1024



Since ‘choosing a coin’ is the ‘step’ and there are 5 coins, there are 5 steps.

By the multiplication rule,

P. 12



Let us study a more complicated example.

The figure shows a road network connecting 4 cities A, B, C and D.

Helen drives her car from city A to city D. How many ways are there for her to complete the journey?

She can choose a route from A to D via B or a route from A to D via C. In this case, we need to apply the addition rule and the multiplication rule together.

By the multiplication rule, the number of ways from A to D via B 3 3 9the number of ways from A to D via C 3 2 6

By the addition rule, the total number of ways from A to D 9 6

15

P. 13

Example 19.5T

Solution:

Robert and Alice are dining in a Chinese restaurant. They decide to order a set meal. Out of 3 kinds of food: steamed, stirfried and deepfried, 2 can be chosen. For steamed food, there are 8 choices. For stirfried food, there are 12 choices. For deepfried food, there are 9 choices. How many ways for them to order the set meal?

There are three combinations: 1. Steamed food and stirfried food;2. Steamed food and deepfried food;3. Stirfried food and deepfried food.

Therefore, by the addition rule and the multiplication rule, the number of choices

91298128 276



P. 14

The product of the first n natural numbers is denoted by n!, which is read as ‘n factorial’.

19.19.22 PermutationPermutation

AA. . Factorial NotationFactorial Notation

For example, 4! 4 3 2 1 120

Notice that 5! 5 4!.

In general, we have:

n! n (n – 1) (n – 2) … 2 1

n! n (n – 1)!

5! 5 4 3 2 1 720

We can press the function key x! or n! on a calculator to find the value of n factorial.

P. 15

We can express the product of some integers in terms of factorial notation.


AA. . Factorial NotationFactorial Notation

For example:

7 6 5 4123

1234567

!3

!7

2 4 6 8 10 12 14

27 (1 2 3 4 5 6 7) 27 7!

P. 16

The letters A, B and C can be arranged as:ABC, ACB, BAC, BCA, CAB, CBA


BB. . Concept and Notation of PermutationConcept and Notation of Permutation

Each of the above arrangements, such as ABC, is called a permutation.

A permutation of n objects is an arrangement of the objects in a definite order.

In the above example, each arrangement of the letters A, B, C consists of a complete list of the letters without repetition.

So there are six permutations of these letters.

P. 17

Consider arranging 4 letters A, B, C and D in a line.



The number of permutations of n distinct objects without repetition is n (n 1) … 2 1 n!.

Position 1 Position 2 Position 3 Position 4

We can arrange the letters using the following 4 steps. (a) Consider the above figure. Select a letter and write it down in

position 1. This can be done in 4 ways.(b) After selecting the letter in position 1, there are 3 letters left.

For position 2, a letter can be selected in 3 ways.(c) After selecting the letters in positions 1 and 2, there are 2 letters left.

For position 3, a letter can be selected in 2 ways.(d) For the final position, a letter can be selected in 1 way.

By the multiplication rule in the counting principle, the number of ways to arrange 4 different letters into a line 4 3 2 1 4!.

P. 18

Example 19.6T

Solution:


A four-digit number is formed with the digits ‘3, 5, 7, 8’ where each digit can be used only once in a number. (a) How many different numbers can be formed? (b) List all the even numbers formed.


(a) Number of four-digit numbers !424

(b)

The even numbers formed are: 3578, 3758, 5378, 5738, 7358, 7538.

P. 19



Sometimes, we need to take out r objects from n distinct objects and arrange them in order.We regard two arrangements as different if the order of the objects is different.In this case, each arrangement is called a permutation of r objects selected from n objects.

Position 1 Position 2 ... Position (r – 1) Position r

Consider the following figure.

For position 1, n objects can be selected. For position 2, (n – 1) objects can be selected.

For position r, (n – r 1) objects can be selected. After the first (r 1) selections, the number of objects left is n (r 1) n r 1.

...

P. 20


Position 1 Position 2 ... Position (r – 1) Position r

By the multiplication rule, the total number of permutations n (n – 1) (n – 2) ... (n – r 1)

1...)(

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

rn

rnrnnnn

n objects (n – 1) objects (n – r 2) objects (n – r 1) objects

)!(

!

rn

n


n

which is denoted by Pr , where n and r are positive integers.

The number of permutations of r objects from n distinct objects

without repetition )!(

!

rn

nPn

r

P. 21



!0

!

)!(

! n

nn

nPn

n

Notes:From the formula , when r n, is just the number of

permutations of n objects:

nr

nr P

rn

nP

)!(

!

!1

!

n

nPn

n

Note that 0! have not been defined. In order to be consistent, we define

0! 1. Thus,

which is the same as the previous result.We can use the function key nPr on a calculator to find the value of Pr

n.

P. 22

Example 19.7TGiven six digits: 2, 4, 5, 6, 8, 9. How many different three-digit numbers can be formed if each digit can be used only once?

Solution:



Number of possible outcomes 6

3P120

P. 23

Suppose Andy, Bobby and Chris are members of the class committee. A chairperson, a secretary and a treasurer are to be selected from them. How many possible results are there?


CC. . Applications of PermutationApplications of Permutation

By listing all of the possibilities, we find that there are 6 different possible results.

However, it is time-consuming to list all of the possibilities if more people are involved.

Chairperson Secretary Treasurer

Andy Bobby Chris

Andy Chris Bobby

Bobby Andy Chris

Bobby Chris Andy

Chris Andy Bobby

Chris Bobby Andy

Therefore, the number of possible results 3! 6, which is the same as we found by listing.

If we consider the chairperson as position 1, the secretary as position 2 and the treasurer as position 3, the problem can beregarded as a permutation of 3 objects.

P. 24

Example 19.8T5 girls and 3 boys are selected as members of the school debate team which is comprised of a first speaker, a second speaker and a third speaker. Suppose Lily is the first speaker. Find the number of different teams that can be formed.

Solution:



2 speakers are left to be selected from 7 students.Number different teams

72P

42

P. 25

Example 19.9T

Solution:

The Chan family is having a photo taken at the entrance to Disneyland. The family consists of Mr. Chan, Mrs. Chan and their three children. Find the number of ways they could line up if (a) Mr. and Mrs. Chan must stand at the two sides; (b) Mr. and Mrs. Chan must stand next to each other; (c) Mr. Chan stands in the middle and Mrs. Chan stands next to him.



(a) As the positions of Mr. and Mrs. Chan are fixed, it is necessary to arrange the children only.

Number of ways 2 3! 12

(b) As Mr. and Mrs. Chan must stand next to each other, we can regard them as one unit. Therefore, we have (5 – 1)! arrangements. Moreover, Mr. and Mrs. Chan can stand in 2 ways.

By the multiplication rule, the number of ways (5 – 1)! 2! 48

Moreover, Mr. and Mrs. Chan can stand in 2 ways.

P. 26

Example 19.9T

Solution:

The Chan family is having a photo taken at the entrance to Disneyland. The family consists of Mr. Chan, Mrs. Chan and their three children. Find the number of ways they could line up if (a) Mr. and Mrs. Chan must stand at the two sides; (b) Mr. and Mrs. Chan must stand next to each other; (c) Mr. Chan stands in the middle and Mrs. Chan stands next to him.



(c) As Mr. and Mrs. Chan must stand next to each other and the position are fixed, it is necessary to arrange the children only.

Number of ways 2 3! 12Moreover, Mr. and Mrs. Chan can stand in 2 ways.

P. 27

Combination concerns the number of ways to choose r objects from n objects, but unlike permutation, the order of the r objects is not considered.

19.19.33 CombinationCombination

Suppose we want to choose 3 distinct letters are chosen from 5 letters A, B, C, D and E.

Since we do not concern about the order of the letters, the following 6 permutations are regarded as the same selection: {A, B, C}.

{A, B, C} , {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, {C, B, A}

AA. . Concept and Notation of CombinationConcept and Notation of Combination

We can choose {A, B, C}, {B, C, E}, {C, D, E} and so on.

Such a selection is called a combination.

A combination is a selection of certain objects without considering the order.

P. 28

We know that the number of permutations of 3 letters from55 letters is P3.



To find the number of combinations:

Step 1:Select any 3 letters from the 5 letters. Let the number of ways be N.

Step 2:Arrange the 3 letters selected in Step 1. The number of ways is 3!.

By the multiplication rule in the counting principle,5

3!3 PN

10!3

53

P

N

We denote by .53

53

!3C

P

P. 29

In general, we denote

where n and r are positive integers.



The number of combinations of r objects from n distinct objects

without repetition !)!(

!

!

!

rrn

n

r

PC n

r

!)!(

!

!

!

rrn

n

r

PC n

r

!2)!25(

!552

C

)12()123(

12345

12

45

For example,

10

We can use the function key nCr on a calculator to find the value of Cr

n.

P. 30

Example 19.10T

Solution:

Three different numbers are selected from {1, 4, 5, 7, 9}. (a) How many different combinations of the numbers are there? (b) List all the combinations that include ‘5’.



(a) Total number 5 Number to be selected 3 Number of combinations 5

3C

(b) Possible combinations are:{1, 4, 5}, {1, 7, 5}, {1, 9, 5}, {4, 7, 5}, {4, 9, 5}, {7, 9, 5}.

10

P. 31


Let us see some of applications of combination.

BB. . Applications of CombinationApplications of Combination

P. 32

Example 19.11T

Solution:

In a class of 30 students, 6 are selected to be the class representatives. How many ways are there to select the representatives if (a) both the monitor and monitress are selected? (b) both the monitor and monitress are not selected?



(a) Total number of students 30 – 2

Number of students to be selected 6 – 2

284C475 20

286C

740 376

28

4The total number of ways

(b) Total number of students 30 – 2 28

The total number of ways

P. 33



In solving counting problems, it is important to identify whether permutation or combination is involved.

1. For problems that involve ordering, Prn should be used.

2. For problems in which ordering is not important, Crn should be used.

P. 34

Example 19.12TA research team consists of 10 boys and 6 girls. 3 of them are selected to present their studies to a class. (a) In how many ways can the team be formed? (b) If the team consists of 2 boys and 1 girl, in how many ways can the

team be formed? (c) If the team consists of 1 boy and 2 girls, in how many ways can the

team be formed? Solution:



(a) The total number of ways 163C 560

(b) Number of ways to select 2 boys 45102 C

Number of ways to select 1 girl 661 C

Total number of ways 645 270

(c) Number of ways to select 1 boy 10101 C

Number of ways to select 2 girls 1562 C

Total number of ways 1510 150

P. 35

19.1 Counting Principle

Chapter Chapter SummarySummary

1. Addition Rule in the Counting PrincipleThere are k different choices to finish a task and there are n1 ways to finish the 1st choice, n2 ways to finish the 2nd choice, ... , andnk ways to finish the kth choice, then the total number of ways to finish the task n1 n2 ... nk.

2. Multiplication Rule in the Counting PrincipleSuppose a task can be divided into k steps and each step will not affect the others. If there are n1 ways to finish the 1st step, n2 ways to finish the 2nd step, ... , and nk ways to finish the kth step, then the total number of ways to finish the task n1 n2 … nk.

P. 36

1. A permutation of n objects is an arrangement of the objects in a definite order.

Chapter Chapter SummarySummary

2. The number if permutations of n distinct objects without repetition is n! n (n 1) (n 2) … 2 1

19.2 Permutation

3. A permutation of r objects selected from n objects is an arrangement of r objects from n distinct objects.

4. The number of permutation of r objects from n distinct objects without repetition

nrP

)!(

!

rn

n

P. 37

1. A combination is a selection of certain objects without considering the order.

Chapter Chapter SummarySummary19.3 Combination

2. The number of combinations of r objects from n distinct objects without repetition

!)!(

!!

!

rrn

nr

PC n

r

19 19.2permutation 19.3combination chapter summary case study permutation and combination...

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