19 Permutation and Combination Case Study 19.1 Counting Principle Chapter Summary

Case Study Andy joins a summer camp. If each team member has to shake hands with all other members once, then how many times does each team need to shake hands? You may try to work it out by counting but it won’t be easy. 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. If there are 4 members in each team, what is the total number of times each team must 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 CD Therefore, the total number of times each team must shake hands is 6.

19.1 Counting Principle A. Addition Rule in the Counting Principle 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. 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. Therefore the total number of ferry trips that he can choose from is 32  28  60 ferry tips from company A ferry tips from company B

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

Example 19.1T 19.1 Counting Principle Solution: A. Addition Rule in the Counting Principle Example 19.1T 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? Solution: 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 have

19.1 Counting Principle B. Multiplication Rule in the Counting Principle Suppose there is only one task. If the procedure for finishing the task 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 is 4  3  12 number of ways of choosing a filling from the 4 choices number of ways of choosing a drink from the 3 choices

19.1 Counting Principle B. Multiplication Rule in the Counting Principle 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.

19.1 Counting Principle B. Multiplication Rule in the Counting Principle 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: k  3 Step 1: Choose a literary book from 10 choices Step 2: Choose a fiction book from 15 choices Step 3: Choose a science book from 12 choices n1  10 n2  15 n3  12 Therefore, the number of choices is 10  15  12  1800.

Example 19.2T 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle 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 The first step is to choose a day and the second step is to choose a match.

Example 19.3T 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle 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

Example 19.4T 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Example 19.4T 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? Solution: We can interpret the ‘step’ as ‘choosing a coin’. By the multiplication rule, the number of ways of putting the coins into the 4 boxes Since ‘choosing a coin’ is the ‘step’ and there are 5 coins, there are 5 steps.

19.1 Counting Principle B. Multiplication Rule in the Counting Principle 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  9 the 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

Example 19.5T 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Example 19.5T 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? Solution: 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

19.2 Permutation A. Factorial Notation The product of the first n natural numbers is denoted by n!, which is read as ‘n factorial’. n!  n  (n – 1)  (n – 2)  …  2  1 For example, 4!  4  3  2  1  120 We can press the function key x! or n! on a calculator to find the value of n factorial. 5!  5  4  3  2  1  720 Notice that 5!  5  4!. In general, we have: n!  n  (n – 1)!

19.2 Permutation A. Factorial Notation We can express the product of some integers in terms of factorial notation. For example:  7  6  5  4  2  4  6  8  10  12  14  27  (1  2  3  4  5  6  7)  27  7!

19.2 Permutation B. Concept and Notation of Permutation The letters A, B and C can be arranged as: ABC, ACB, BAC, BCA, CAB, CBA 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.

Position 1 Position 2 Position 3 Position 4 19.2 Permutation B. Concept and Notation of Permutation Consider arranging 4 letters A, B, C and D in a line. 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!. The number of permutations of n distinct objects without repetition is n  (n  1)  …  2  1  n!.

Example 19.6T 19.2 Permutation Solution: B. Concept and Notation of Permutation Example 19.6T 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. Solution: (a) Number of four-digit numbers (b)  The even numbers formed are: 3578, 3758, 5378, 5738, 7358, 7538.

19.2 Permutation B. Concept and Notation of Permutation 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. Consider the following figure. Position 1 Position 2 ... Position (r – 1) Position r For position 1, n objects can be selected. For position 2, (n – 1) objects can be selected. After the first (r  1) selections, the number of objects left is n  (r  1)  n  r  1. ... For position r, (n – r  1) objects can be selected.

19.2 Permutation B. Concept and Notation of Permutation By the multiplication rule, the total number of permutations  n  (n – 1)  (n – 2)  ...  (n – r  1) n which is denoted by Pr , where n and r are positive integers. Position 1 Position 2 ... Position (r – 1) Position r n objects (n – 1) objects (n – r  2) objects (n – r  1) objects The number of permutations of r objects from n distinct objects without repetition 

19.2 Permutation B. Concept and Notation of Permutation Notes: From the formula , when r  n, is just the number of permutations of n objects: 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 Prn.

Example 19.7T 19.2 Permutation Solution: B. Concept and Notation of Permutation Example 19.7T Given 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

19.2 Permutation C. Applications of Permutation 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? By listing all of the possibilities, we find that there are 6 different possible results. Chairperson Secretary Treasurer Andy Bobby Chris However, it is time-consuming to list all of the possibilities if more people are involved. If we consider the chairperson as position 1, the secretary as position 2 and the treasurer as position 3, the problem can be regarded as a permutation of 3 objects. Therefore, the number of possible results  3!  6, which is the same as we found by listing.

Example 19.8T 19.2 Permutation Solution: C. Applications of Permutation Example 19.8T 5 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

Example 19.9T 19.2 Permutation Solution: C. Applications of Permutation Example 19.9T 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 Mr. and Mrs. Chan must stand at the two sides; Mr. and Mrs. Chan must stand next to each other; Mr. Chan stands in the middle and Mrs. Chan stands next to him. Solution: As the positions of Mr. and Mrs. Chan are fixed, it is necessary to arrange the children only. Moreover, Mr. and Mrs. Chan can stand in 2 ways. Number of ways  2  3! (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!

Example 19.9T 19.2 Permutation Solution: C. Applications of Permutation Example 19.9T 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 Mr. and Mrs. Chan must stand at the two sides; Mr. and Mrs. Chan must stand next to each other; Mr. Chan stands in the middle and Mrs. Chan stands next to him. Solution: (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. Moreover, Mr. and Mrs. Chan can stand in 2 ways. Number of ways  2  3!

19.3 Combination A. Concept and Notation of Combination 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. Suppose we want to choose 3 distinct letters are chosen from 5 letters A, B, C, D and E. We can choose {A, B, C}, {B, C, E}, {C, D, E} and so on. 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} Such a selection is called a combination. A combination is a selection of certain objects without considering the order.

19.3 Combination A. Concept and Notation of Combination We know that the number of permutations of 3 letters from 5 5 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, We denote by .

19.3 Combination A. Concept and Notation of Combination In general, we denote where n and r are positive integers. The number of combinations of r objects from n distinct objects without repetition  For example, We can use the function key nCr on a calculator to find the value of Crn.

Example 19.10T 19.3 Combination Solution: A. Concept and Notation of Combination Example 19.10T 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’. Solution: (a) Total number  5 Number to be selected  3 Number of combinations (b) Possible combinations are: {1, 4, 5}, {1, 7, 5}, {1, 9, 5}, {4, 7, 5}, {4, 9, 5}, {7, 9, 5}.

19.3 Combination B. Applications of Combination Let us see some of applications of combination.

Example 19.11T 19.3 Combination Solution: B. Applications of Combination Example 19.11T 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? Solution: (a) Total number of students  30 – 2  28 Number of students to be selected  6 – 2  4 The total number of ways (b) Total number of students  30 – 2  28 The total number of ways

19.3 Combination B. Applications of Combination 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.

Example 19.12T 19.3 Combination Solution: B. Applications of Combination Example 19.12T A 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 (b) Number of ways to select 2 boys Number of ways to select 1 girl  Total number of ways (c) Number of ways to select 1 boy Number of ways to select 2 girls  Total number of ways

Chapter Summary 19.1 Counting Principle 1. Addition Rule in the Counting Principle 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. 2. 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.

Chapter Summary 19.2 Permutation 1. A permutation of n objects is an arrangement of the objects in a definite order. The number if permutations of n distinct objects without repetition is  n!  n  (n  1)  (n  2)  …  2  1 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

Chapter Summary 19.3 Combination 1. A combination is a selection of certain objects without considering the order. 2. The number of combinations of r objects from n distinct objects without repetition

Follow-up 19.1 19.1 Counting Principle Solution: A. Addition Rule in the Counting Principle Follow-up 19.1 An insurance company conference will be held in a leap year on a day either in January or February. How many different choices are there for the day of the conference? Solution: If the company chooses a day in January, then the company has 31 choices. If the company chooses a day in February, then the company has 29 choices. The number of choices the company has  31 + 29

Follow-up 19.2 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Follow-up 19.2 Helen has 3 pairs of trousers and 4 T-shirts to choose from. How many ways can she dress for her meeting with friends? Solution: By the multiplication rule, the number of ways The first step is to choose a pair of trousers and the second step is to choose a T-shirt.  3  4

Follow-up 19.3 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Follow-up 19.3 The cabin numbers on a cruise ship are formed by a letter followed by a two-digit number from 01 to 99. (a) How many different cabin numbers can be formed? (b) If the letters O and X are not used, how many different cabin numbers can be formed? Solution: (a) By the multiplication rule, the number of different cabin numbers  26  99 (b) Since O and X are not used, there are only 24 letters that can be chosen. Number of choices  24  99

Follow-up 19.4 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Follow-up 19.4 5 students are assigned to 3 rooms. If each room can hold 5 students at most, what is the number of possible ways to assign the students? Solution: We can interpret the ‘step’ as ‘choosing a student’. By the multiplication rule, the number of ways to assign the students Since ‘choosing a student’ is the ‘step’ and there are 5 students, there are 5 steps.  3  3  3  3  3

Follow-up 19.5 19.1 Counting Principle Solution: B. Multiplication Rule in the Counting Principle Follow-up 19.5 Andy is going to join his school sports day. There are 3 sports categories: field events, long-distance races and sprints. He plans to compete in 2 events, each from a different category. There are 8 choices of field events, 4 choices of long-distance races and 3 choices of sprints. How many different choices does Andy have? Solution: There are three combinations: 1. Field events and long-distance races; 2. Field events and sprints; 3. Long-distance races and sprints. Therefore, by the addition rule and the multiplication rule, the number of choices  8  4 + 8  3 + 4  3

Follow-up 19.6 19.2 Permutation Solution: B. Concept and Notation of Permutation Follow-up 19.6 Three letters ‘A, M, E’ are arranged in a row. (a) How many different ways are there to arrange the letters? (b) List all the possible outcomes. Solution: (a) Number of ways to arrange the letters  3! (b)  The possible outcomes are: AME, AEM, MAE, MEA, EAM, EMA.

Follow-up 19.7 19.2 Permutation Solution: B. Concept and Notation of Permutation Follow-up 19.7 Three different letters are selected from the word ‘FORMULA’ and arranged in order. In how may different ways can the selected letters be arranged? Solution: Number of ways

Follow-up 19.8 19.2 Permutation Solution: C. Applications of Permutation Follow-up 19.8 4 students are to be elected from 10 students to 4 different posts in the class committee. (a) Find the number of possible results of the election. (b) Given that one of the 10 students, Winnie, is elected as a member of the committee, find the number of possible results of the election. Solution: (a) Number of possible results (b) 3 posts are left to be elected from 9 students. Number of possible results

Follow-up 19.9 19.2 Permutation Solution: C. Applications of Permutation Follow-up 19.9 An awards ceremony is held at City Hall. The following figure shows the seating reserved for the 10 prizewinners: 5 men and 5 women. Tommy and Amy are two of the prizewinners. Find the number of arrangements for the prizewinners if (a) there are no restrictions; (b) the men and the women must sit in 2 separate rows; (c) Tommy must sit on seat 10 and Amy must sit on seat 5. Solution: (a) Number of arrangements (b) We first arrange the men in a row. The number of ways  5! Then we arrange the women in another row. The number of ways  5! Since there are 2 rows, number of arrangements

Follow-up 19.9 19.2 Permutation Solution: C. Applications of Permutation Follow-up 19.9 An awards ceremony is held at City Hall. The following figure shows the seating reserved for the 10 prizewinners: 5 men and 5 women. Tommy and Amy are two of the prizewinners. Find the number of arrangements for the prizewinners if (a) there are no restrictions; (b) the men and the women must sit in 2 separate rows; (c) Tommy must sit on seat 10 and Amy must sit on seat 5. Solution: (c) As Tommy and Amy have already chosen their seats, we have (10 – 2)! arrangements. Number of arrangements

Follow-up 19.10 19.3 Combination Solution: A. Concept and Notation of Combination Follow-up 19.10 Two different numbers are selected from {0, 2, 5, 7, 9}. (a) How many different combinations of the numbers are there? (b) List all the possible combinations of the numbers. Solution: (a) Total number  5 Number to be selected  2 Number of combinations (b) Possible combinations are: {0, 2}, {0, 5}, {0, 7}, {0, 9}, {2, 5}, {2, 7}, {2, 9}, {5, 7}, {5, 9}, {7, 9}.

Follow-up 19.11 19.3 Combination Solution: B. Applications of Combination Follow-up 19.11 8 boys are trained for the school athletics meet. Mr. Lee will select 4 of them to form a relay team. (a) In how many possible ways can the relay team be formed? (b) Simon hurts his leg, so Mr. Lee will not select him. In how many possible ways can the team be formed now? Solution: (a) Total number of students  8 The total number of ways (b) Total number of students  8 – 1  7 The total number of ways

Follow-up 19.12 19.3 Combination Solution: B. Applications of Combination Follow-up 19.12 8 girls and 6 boys are nominated for class representatives. (a) If 5 students are to be selected, how many different ways are there? (b) If 2 boys and 3 girls are to be selected, how many different ways are there? Solution: (a) Total number of students  8 + 6  14 Total number of ways (b) Number of ways to select 2 boys Number of ways to select 3 girls  Total number of ways