MIND ACTION SERIES THE COUNTING PRINCIPLE AND PROBABILITY GRADE

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1 MIND ACTION SERIES THE COUNTING PRINCIPLE AND PROBABILITY GRADE 12 MARK PHILLIPS

2 THE COUNTING PRINCIPLE AND PROBABILITY GRADE The basic product rule of choices: a1 a2 a3... an 2. The product rule of limited choices: (n objects taken r at a time) 3. Working with letter arrangements: Letters are all different (1) repetition (exponential form) (2) non-repetition (factorial form) Some letters are identical (non-repetition only) 4. Arrangements of objects in a row 5. Passwords and codes 6. Link to Probability 1

3 LESSON 1 (The basic product rule of choices) Example 1 A meal can be made up as follows: Choice 1: Choice 2: meat, fish or chicken mash, chips, baked potato, rice or vegetables How many different meals can be made using these choices if only one item may be selected from each choice? There are 3 possible choices in Choice 1 followed by 5 possible choices in Choice 2. There are therefore 15 possible meals that can be made: meat, mash fish, mash chicken, mash meat, chips fish, chips chicken, chips meat, baked potato fish, baked potato chicken, baked potato meat, rice fish, rice chicken, rice meat, vegetables fish, vegetables chicken, vegetables According to the fundamental counting principle, there are possible meals that can be made. Example 2 A gift pack can be made up as follows: Choice 1: Choice 2: Choice 3: Choice 4: Choose one CD out of a possible 4 different CD s Choose one packet of chips out of a possible 5 different types Choose one chocolate type out of a possible 12 different types Choose one type of fruit out of a possible 3 different fruit types How many different gift packs can be made? different gift packs 2

4 Example 3 A menu for a social event is given below. A person attending the function must choose only ONE item from each category, that is starters, main course and dessert. MENU STARTERS MAIN COURSE DESSERT Soup Fried Chicken Ice-cream Salad Fillet steak Fruit salad Chicken Curry Vegetable Curry How many different meal combinations can be chosen? A particular person wishes to have soup as his starter and chicken as his main course. How many different meal combinations does he have? meal combinations meal combinations 3

5 LESSON 2 (The product rule of limited choices) Example 1 In how many ways can 6 different people be seated in the first six seats in a movie theatre? In this example, any one of the 6 people can be chosen to sit in the first seat. Thereafter, only one of the remaining 5 people can be chosen to sit in the second seat. After this, only one of the remaining 4 people can be chosen to sit in the third seat. After this, only one of the remaining 3 people can be chosen to sit in the fourth seat. After this, only one of the remaining 2 people can be chosen to sit in the fifth seat. The remaining person will then have to occupy the sixth seat. The number of ways that we can seat these people is therefore as follows: ! 720 ways Example 2 In how many ways can 7 vacant places be filled by 10 different people? 10 different people can occupy 7 places in the following ways: (only 7 places can be occupied) We can write this as follows: ! 10! ! (10 7)! Example 3 There are 12 different singers that are hoping to occupy the first three places in SA s GOT TALENT. In how many different ways can the first three places be occupied? In this example there are 12 people to be arranged in 3 different ways. The number of possible arrangements will be: 12! 12! (12 3)! 9! 4

6 Example 4 In how many ways can 1000 vacant places be filled by 80 different people? 1000! ( )! (Here is where the rule n! ( n r)! is useful) Example 5 In how many different ways can the letters of the word tirade be arranged in any order without repetition to form a four-letter word, chosen at random, such that each word will not contain the letter d? Without d : 5! (5 4)! 5

7 LESSON 3A (Working with letter arrangements) Letter arrangements where all letters are different Example 1 Consider the word PARKTOWN (all letters are different). You are required to form different eight-letter word arrangements using the letters of the word PARKTOWN. An example of a word arrangement would be the word APKROTWN. This arrangement of the letters need not make any sense. How many possible word arrangements can be made if: the letters may be repeated? the letters may not be repeated? An example of an eight-letter word arrangement where the letters may be repeated is AAPKWANO. This means that for the first letter of the word arrangement, any of the eight letters in the word PARKTOWN can be used. For the second letter, any of the eight letters may be used again. In the word arrangement AAPKWANO, the letter A is used as the first letter and then again as the second letter as well as the sixth letter. In forming a word arrangement, there are 8 possible letters that can be used for the first letter. For the second letter, we can still use 8 letters (repeating the letters is allowed). For the third letter there are still 8 possible letters available to use. From the fundamental counting principle, there are: word arrangements. An example of an eight-letter word arrangement where the letters may not be repeated is NOTWARPK. This means that for the first letter of the word arrangement, any of the eight letters in the word PARKTOWN can be used. However, for the second letter, only seven letters may be used since the first letter may not be used again. In the word arrangement NOTWARPK, the letter N is used as the first letter but not again as the second letter. In forming a word arrangement, there are 8 possible letters that can be used for the first letter. For the second letter, we can use 7 letters (repeating the letters is not allowed). For the third letter, there will be 6 possible letters available to use. For the eighth letter, there will only be one choice. This will be the last remaining letter that was not used. From the fundamental counting principle, there are: ! word arrangements. 6

8 Example 2 Consider the word LOVERS. (d) How many six-letter word arrangements can be made if the letters may be repeated? How many six-letter word arrangements can be made if the letters may not be repeated? How many four-letter word arrangements can be made if the letters may be repeated? How many four-letter word arrangements can be made if the letters may not be repeated? When the letters may be repeated, we use exponential notation: When letters may not be repeated, we use factorial notation: ! 720 (d) ! (6 4)! 7

9 LESSON 3B (Working with letter arrangements) Letter arrangements where some letters are identical Consider the letters of the word DAD. Suppose that all of the letters are treated as being different Let the first D be D 1 and the last D be D 2. The word arrangements are: DA 1 D 2 D2 AD 1 AD 1 D 2 AD 2 D 1 D1 D 2 A D2 DA 1 There are 3 letters in the word DAD. There are therefore 3! possible word arrangements if the two D s are treated as being different letters. But suppose that the D s are identical. Then we can remove the subscripts and obtain the following word arrangements: DAD DAD ADD ADD DDA DDA There are three arrangements now: DAD ADD DDA This means that 6 arrangements ! 3 arrangements ! There are 3 letters in the word DAD. There are therefore 3! possible word arrangements if the two D s are treated as identical 2! letters. In the number 3!, the numerator represents the total number of letters in the word DAD, 2! which is 3. The denominator represents the fact that the letter D is written twice in the word. Example Consider the letters of the word NEEDED. How many word arrangements can be made with this word? How many word arrangements can be made with this word if the word starts and ends with the same letter? The total possible word arrangements (repeated letters are treated as identical) is: 6! ! 2! The numerator represents the 6 letters in the word NEEDED. The denominator represents the three E s and the two D s. 8

10 The only possibilities with the word NEEDED if you start and end with the same letter are: Option 1 Option 2 D D E E With the first option the letters in between the two D s will be NEEE. The possible word arrangements will then be: 4! ! The numerator represents the 4 letters in the word NEEE (ignore the two D s). The denominator represents the three E s. With the second option E the letters in between the two E s will be NDED. The possible word arrangements will then be: 4! ! 2 1 The numerator represents the 4 letters in the word NDED (ignore the two E s). The denominator represents the two D s. The total number of possible word arrangements that can be made if the word starts and ends with the same letter will therefore be: 4! 4! ! 2! 9

11 LESSON 4 (Arrangements of objects in a row) Example 1 (d) (e) (f) In how many ways can three boys and two girls sit in a row? In how many ways can they sit in a row if a boy and his girlfriend must sit together? In how many ways can they sit in a row if the boys and girls are each to sit together? In how many ways can they sit in a row if just the girls are to sit together? In how many ways can they sit in a row if just the boys are to sit together? In how many ways can they sit in a row if the boys and girls are to alternate? The five persons can sit in a row in ! 120 ways. The boy can sit to the left or right of his girlfriend: BG or GB Either the boy or girl could sit in the left seat. If it is the boy, then the girl will sit in the right seat. If the girl sits in the left seat, then the boy will sit in the right seat. From the counting principle, we can write this as 2 1 2! ways of sitting. Now let s treat the couple (BG or GB) as a single entity. This means that we have four objects : one couple, two boys and one girl. These four objects can be seated in 4! ways: couple (4 choices) couple (3 choices) couple (2 choices) couple (1 choice) From the counting principle, the final answer is: ways that the couple can sit together ways that the couple sits with the other 3 2! 4! 48 ways (d) There are two ways to arrange them: BBBGG or GGBBB. In each case, the boys can sit in 3! ways and the girls can sit in 2! ways. Therefore, they can together sit in 2 3! 2! 24 ways. There are 4 ways of arranging them: GGBBB or BGGBB or BBGGB or BBBGG. There are 2! ways for the girls to sit together. The girls as an entity can sit with the other 3 boys in 4! ways. Two girls Two girls Two girls Two girls The final answer is therefore: 2! 4! 48 ways 10

12 (e) There are 3 ways of distributing them: GGBBB or GBBBG or BBBGG. There are 3! ways for the boys to sit together. The boys as an entity can sit with the girls in 3! different ways. Three boys Three boys Three boys The final answer is therefore: 3! 3! 36 ways (f) There is only one way that this can happen in this case: BGBGB (not GBGBB). Example 2 There are ways ways. The letters of the word LOVERS are to be used to form other word arrangements without repeating the letters. How many word arrangements are possible if the letters L and O are NOT to be next to each other? If letters are in any order: 6! If L and O are together: 2!5! If L and O are not together: 6! 2!5! 480 Example 3 Consider the word MATHEMATICS In how many unique ways can the letters from the word be arranged? The letter H and E cannot be placed next to each other. Find the number of arrangements in this case. 11! !2!2! If H and E together: If H and E not together: 2!10! 2!2!2! 11! 2!10! !2!2! 2!2!2! 11

13 Example 4 Alan and Jason enter a race. There are 10 competitors and 10 lanes. Calculate the number of possible arrangements in the starting line if: Alan and Jason are place next to each other. Alan is placed in the first lane and Jason is NOT place next to Alan. 9! 2! ! Example 5 Peter, Simon and Thabo are to run in a race which has 8 runners in total. What is the total number of arrangements at the start? If Peter, Simon and Thabo are to be next to each other in any order, how many arrangements are possible? Peter is in Lane 1, Simon is in Lane 2 and Thabo is in Lane 3, how many arrangements are possible? 8! ! 3! ! 120 Example 6 A group of people plan a trip to Europe and want to visit Germany, France, Italy, Sweden, Spain and Norway. The order of visits is chosen randomly. (d) If Germany, France and Italy are visited one after the other in that order, find the number of different orders of their visits. If Germany, France and Italy must be visited one after the other in any order, find the number of different orders of their visits. If Germany, France and Italy must be visited first in any order, find the number of different orders of their visits. If Germany, France and Italy must be visited first in that order, find the number of different orders of their visits. 1 4! 24 3! 4! ! 3! 36 (d) 1 1 3! 6 12

14 Example 7 There are 7 different shirts and 4 different pairs of trousers in a cupboard. The clothes have to be hung on a rail. (d) In how many different ways can the clothes be hung on the rail? In how many different ways can the clothes be arranged if all the shirts are to be hung next to each other and the pairs of trousers are to be hung next to each other on the rail? Suppose that any pair of trousers will hang at the beginning of the rail and a shirt will hang at the end of the rail. In how many different ways can this happen? Suppose that a pre-selected pair of trousers must hang at the beginning and a preselected shirt must hang at the end of the rail. In how many different ways can this happen? 11! ! 7! 4! ! (d) 1 9!

15 LESSON 5 (Passwords and codes) Example 1 A combination to a lock is formed using three letters of the alphabet, excluding the letters O, Q, S, U, V and W and using any three digits. The numbers and letters can be repeated. Calculate the possible number of combinations, chosen at random, if the combination: starts with the letter X and ends with the number 6. has exactly one X. has one or more number 6 in it ( ) ( ) ( ) Method 1 Total number of possible combinations: Number of combinations without 6: Number of combinations with at least one 6: Method 2 One 6: Two 6 s: Three 6 s:

16 Example 2 Consider the digits 1,2,3,4,5,6,7,8 How many 2-digit numbers can be formed if repetition is allowed? How many 4-digit numbers can be formed if repetition is not allowed? How many numbers between 4000 and 5000 can be formed? ! (8 4)! Example 3 Passwords must be formed from the letters of the alphabet followed by digits 0 to 9. Letters and digits may be repeated. Calculate the number of passwords that can be created using 2 letters followed by 2 digits. Suppose that different passwords are needed. The two letters will be used but more digits are needed. Calculate the least number of digits needed n n n log n 3, digits are needed 15

17 LESSON 6 Link to probability Example 1 What is the probability that if the letters of the word tirade are arranged in any order without repetition, a four-letter word, chosen at random, will not contain the letter d? Without d : With d : Probability: 5! (5 4)! 6! (6 4)! Example 2 Consider the letters of the word DREAMS. If the letters are arranged in any order without repetition to form different words, what is the probability that the word formed will start with D and end with S? The number of ways that the six letters can be arranged is 6! Let event E be defined as the event that the word formed will start with D and end with S. The number of arrangement for event E is 4! The probability of event E happening is therefore: 4! 4! 1 6! 6 5 4! 30 Example 3 The letters of the word LOVERS are to be used to form other word arrangements without repeating the letters. What is the probability that the letters L and O are NOT to be next to each other? If letters are in any order: 6! If L and O are together: 2!5! If L and O are not together: 6! 2!5! 480 Probability: 6! 2!5! 2 6! 3 16

18 Example 4 Consider the letters of the word MATHEMATICIAN. The repeated letters are identical. If the letters are arranged in any order without repetition to form different words, what is the probability that the word formed will: start and end with M? end with the letter N? start and end with M? M A T H E A T I C I A N M 11! 3! 2! 2! 13! 3!2!2!2! 1 78 end with the letter N? M A T H E M A T I C I A N 12! 3!2!2! 13! 3!2!2!2! 2 13 Example 5 Peter, Simon and Thabo are to run in a race which has 8 runners in total. What is the total number of arrangements at the start? Find the probability that Peter, Simon and Thabo are to be next to each other in any order. Find the probability that Peter is in Lane 1, Simon is in Lane 2 and Thabo is in Lane 3. 8! ! 3! 3 8! ! 1 8!

19 Example 6 A group of people plan a trip to Europe and want to visit Germany, France, Italy, Sweden, Spain and Norway. The order of visits is chosen randomly. If Germany, France and Italy are grouped together in that order, find the number of different orders of their visits. What is the probability that they will visit Germany, France and Italy one after the other in any order? 1 4! 24 3! 4! 1 6! 5 Example 7 There are 7 different shirts and 4 different pairs of trousers in a cupboard. The clothes have to be hung on a rail. In how many different ways can the clothes be hung on the rail? In how many different ways can the clothes be arranged if all the shirts are to be hung next to each other and the pairs of trousers are to be hung next to each other on the rail? What is the probability that a pair of trousers will hang at the beginning of the rail and a shirt will hang at the end of the rail? 11! ! 7! 4! ! ! 55 Example 8 A combination to a lock is formed using three letters of the alphabet, excluding the letters O, Q, S, U, V and W and using any three digits. The numbers and letters can be repeated. Calculate the probability that a combination, chosen at random: starts with the letter X and ends with the number 6. has exactly one X. has one or more number 6 in it. 18

20 Let A be the event that a number plate starts with the letter X and ends with the number 6. Since 20 letters and 10 digits can be used, the number of plates possible will be: For event A, the number of possibilities is reduced to: Therefore, the probability of event A happening is: Let B be the event of choosing exactly one X. The number of possible ways of event B happening is: ( ) ( ) ( ) Therefore, the probability of event B happening is: Let C be the event of at least one 6 being chosen. Method 1 Total number of possible combinations: Number of combinations without 6: Number of combinations with at least one 6:

21 The probability of at least one 6 is: Method 2 The probability of event C happening can determined by using the fact that P(C) 1 P(not C) P(C) P(C) Method 3 One 6: Two 6 s: Three 6 s: The probability of at least one 6 is:

Introduction. Firstly however we must look at the Fundamental Principle of Counting (sometimes referred to as the multiplication rule) which states:

Introduction. Firstly however we must look at the Fundamental Principle of Counting (sometimes referred to as the multiplication rule) which states: Worksheet 4.11 Counting Section 1 Introduction When looking at situations involving counting it is often not practical to count things individually. Instead techniques have been developed to help us count