1. Quiz: Draw the unit circle: Include: (1)All “nice” angles in degrees (2) All “nice” angles in radians (3) The (x, y) pairs for each point on the unit circle where the terminal side of the “nice” angles intersects the unit circle.

2. Section 9-3 (part 1) Counting

3. HOMEWORK See Blog See Blog(worksheet)

4. What you’ll learn about How to count using: 1. The multiplication principle of counting. 2. Permutations (formula and calculator). 3. Combinations (formula and calculator)

5. HOMEWORK Section 9-3 Section 9-3 See Blog (16 problems)

6. What you’ll learn about What is Probability? Simple Probability by counting: (1 event) Simple Probability using the outcome space: (1 event) Probability of multiple independent events : (>1 event) Probability of multiple dependent events : (>1 event) Conditional Probability (an event depends upon previous event: (>1 event) previous event: (>1 event) Using a binomial expansion to determine the probability of achieving 0,1,2,…,’n’ successes out of ‘n’ trials. of achieving 0,1,2,…,’n’ successes out of ‘n’ trials.

7. Vocabulary: Continuous Data: Data that contains an infinite number of points within a data range. number of points within a data range. Example: number line, there are an infinite number of numbers in the interval [3, 5]. of numbers in the interval [3, 5]. Continuous data does not lend itself very well to “counting.”

8. Vocabulary: Discrete Data: Data sets that contain a finite number of data points. number of data points. Example: The number of ways you can assign students to seats in this class. students to seats in this class. Discrete mathematics deals with counting.

9. Given the first letter above, the second letter could be: Arranging 3 Objects in Order How many ways can you arrange the letters A, B, and C ? A, B, and C ? A B C B or C A or C A or B Any one of the following 3 could be the 1 st letter. The only option for the 3 rd letter in each case is: ABC ACB BAC BCA CBA CAB SIX ways

10. “Tree diagram” How many way can you arrange the letters A, B, and C ? A, B, and C ? A B C B C A C B A C B C A A B ABC,ABC,ABC,ABC, ACB,ACB,ACB,ACB, BAC,BAC,BAC,BAC, BCA,BCA,BCA,BCA, CBA,CBA,CBA,CBA, CABCABCABCAB SIX ways We call this a tree diagram. ABCABCABCABC

11. The “multiplication principle.” When arranging things in order (letters A, B, and C), the total number of possible ways to arrange things is the product of the number of possibilities for each step. of the number of possibilities for each step.

12. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the first number? 10

13. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the 2nd number? 10 * 10

14. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the 3rd number? 10 * 10

15. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the 1 st letter? 10 * 10 * 26

16. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the 2nd letter? 10 * 10 * 26

17. Using the Multiplication Principle If a license plate has four letters followed by three numerical digits. Find the number of different license plates that could be formed if there is no restriction on the letters or digits that can be used. # # # A B C How many possibilities for the 3rd letter? 10 * 10 * 26 = 17,576,000 Wow!

18. Your Turn: 1. How many distinct license plates can be made using 6 digits (numerals 0 – 9)? 6 digits (numerals 0 – 9)? 2. How many distinct license plates can be made using 2 digits (numerals 0 – 9) and 4 letters ( a – z) ? 2 digits (numerals 0 – 9) and 4 letters ( a – z) ? # # L L L L # # L L L L

19. Using the Fundamental counting Principle to count the # of ways to line up 3 people. The first position in line could be either 1 of 3 people 3 ways. The second position in line could be either 1 of 2 people 2 ways. The last position in line could only be the person left over. Total number of ways = 3 * 2 * 1 = 3! “!” means “factorial” Since you “use a person up” (no replacement) each subsequent position has 1 less possibility than the subsequent position has 1 less possibility than the previous position. previous position. 1 way.

20. Effect on Muliplication Principle of counting ( Product of the # of options for each step) arranging without replacement: arranging with replacement: Arranging 3 spaces on a licence plate. Arranging 3 people in a line. Factorial

21. Vocabulary: Factorial: Multiply a positive integer by every positive integer that is smaller than the original integer. integer that is smaller than the original integer. 3! = 3*2*1 3! = 6 Your Turn: 3. Calculate 5! 4. Calculate the total number of distinct license plates that are possible with 3 letters (with replacement) and 3 numbers (without replacement).

22. Using your calculator: Factorial: Select the “math” push button (p/b) Scroll over to “PRB” (probability) Type in the number….7. Now select option 4 (factorial) Hit “enter” Taking the “factorial” of a number can result in HUGE numbers!

23. Your Turn: 5. Use your calculator to find: 8! 40,320

24. Learning to count: How many ways are there to arrange the 13 people in a line? 13 ! 6,227,020,800

25. Your Turn: 6. A car dealership has a large showroom. It has room for 12 cars in a row. How many different ways room for 12 cars in a row. How many different ways can you arrange the 12 cars in order from left to right? can you arrange the 12 cars in order from left to right? 12 ! 479,001,600 7. Use your calculator factorial feature to calculate: 0 ! 0 ! = 1

26. Vocabulary The number of ways a group of items can be arranged is called a permutation. The previous examples were all permutations. Permutations different ways “x” number of items can be arranged in order.

27. Permutations For example: Sean’s band has 10 original songs. The recording company will only accept 4 songs on a demo CD. How many different ways can you choose 4 of the 10 and then arrange them on the demo disk? We call this a permutation of ‘n’ items taken ‘r’ at a time. What if you don’t want to arrange all of the items? Instead you want to pick from a group of items but arrange only a portion of them. How many ways are there to do this? For Sean’s CD: “10 permutate 5”

28. Permutations Counting Formula (We have “n” distinct number of objects to place into “r” number of positions). to place into “r” number of positions). The number of permutations of “n” objects taken “r” at a time, is denoted by: and (We have 10 distinct number of songs to place into 4 number of positions). to place into 4 number of positions).

29. Permutations For example: Sean’s band has 10 original songs. The recoding company will only accept 5 of them on a demo CD. How many different ways can you choose 5 of the 10 and then arrange them on the demo disk?

30. Permutations using your calculator “Math” p/b Scroll to “PRB” Select option “2” “10 permutate 6” Clear your screen then enter “10” Hit “6” then “enter”

31. Your Turn: 8. Calculate “7 permutate 3” on your calculator. 210 9. There are 10 candidates. The one with the highest number of votes will be president, the 2 nd highest will be vice of votes will be president, the 2 nd highest will be vice president and the 3 rd highest will be secretary. president and the 3 rd highest will be secretary. How many ways are there to arrange 3 candidates chosen from a group of 10 in the positions of president, vice president and secretary? 720

32. Permutations. If we were making a permutation using the letters ‘D’, ‘A’, ‘W’, and ‘G’ and ‘D’, ‘A’, ‘W’, and ‘G’ and WADGDAWG would be two distinct words. would be two distinct words. ORDER MATTERS!! (with permutations)  a different order of members is a different group all together!! of members is a different group all together!!

33. Each of these groups is just a permutation of the # of ways to arrange 3 different bills. arrange 3 different bills. 1, 2, 5 1, 5, 2 2, 1, 5 2, 5, 1 5, 1, 2 5, 2, 1 10, 2, 1 10, 1, 2 2, 10, 1 2, 1, 10 1, 10, 2 1, 2, 10 10, 2, 5 10, 5, 2 2, 10, 5 2, 5, 10 5, 10, 2 5, 2, 10 1, 10, 5 1, 5, 10 10, 1, 5 10, 5, 1 5, 1, 10 5, 10, 1 I have 4 bills in my wallet: $1, $2, $5, $10 I have 4 bills in my wallet: $1, $2, $5, $10 How many different sequences of bills can I take out of my wallet, if I only take 3 out? 24 ways

34. I have 4 bills in my wallet: $1, $2, $5, $10 I have 4 bills in my wallet: $1, $2, $5, $10 How many different sums of money can I take out of my wallet, if I only take 3 bills out? 1, 2, 5 1, 5, 2 2, 1, 5 2, 5, 1 5, 1, 2 5, 2, 1 10, 2, 1 10, 1, 2 2, 10, 1 2, 1, 10 1, 10, 2 1, 2, 10 10, 2, 5 10, 5, 2 2, 10, 5 2, 5, 10 5, 10, 2 5, 2, 10 1, 10, 5 1, 5, 10 10, 1, 5 10, 5, 1 5, 1, 10 5, 10, 1 = $8 = $13 = $16 = $17 ORDER Doesn’t MATTER!!  a different order of pulling the same 3 bills out doesn’t make a different sum. of pulling the same 3 bills out doesn’t make a different sum. 4 ways

35. We call this new method of counting a “ combination”. 1, 2, 5 1, 5, 2 2, 1, 5 2, 5, 1 5, 1, 2 5, 2, 1 10, 2, 1 10, 1, 2 2, 10, 1 2, 1, 10 1, 10, 2 1, 2, 10 10, 2, 5 10, 5, 2 2, 10, 5 2, 5, 10 5, 10, 2 5, 2, 10 1, 10, 5 1, 5, 10 10, 1, 5 10, 5, 1 5, 1, 10 5, 10, 1 = $8 = $13 = $16 = $17 ORDER Doesn’t MATTER!!  a different order of pulling the same 3 bills out doesn’t make a different sum. of pulling the same 3 bills out doesn’t make a different sum.

36. Counting the # of ways to arrange choices: Order matters vs. Order Doesn’t matter. ORDER MATTERS!! (with permutations)  a different order of members is a different group all together!! Order matters: “golf” and “flog” are different words using the same 4 letter.. using the same 4 letter.. Order doesn’t matter: when adding the scores of darts tossed at a dart board, getting a ‘3’ first and a ‘5’ second is the same at a dart board, getting a ‘3’ first and a ‘5’ second is the same as getting a ‘5’ first and a ‘3’ second. as getting a ‘5’ first and a ‘3’ second. Since the order of rolling dice doesn’t matter when finding the sum of the two dice we call this a combination. the sum of the two dice we call this a combination.

37. “Order Matters” vs. “Order Doesn’t Matter” Permutation Different order of the same items counted as a separate arrangement  Different ways to line up people/things in order  If you see the words “…in order” in the question  Different presidencies  Different prizes based upon order of finish in a race

38. “Order Matters” vs. “Order Doesn’t Matter” Combination Different order of the same items  can not be counted as separate arrangement  can not be counted as separate arrangement  Different total scores  Different total amounts of money  Different “hands” of dealt card (in games where you can rearrange the cards in your hand once they are dealt) can rearrange the cards in your hand once they are dealt)  Different committees of people

39. Combination: You are paying a for groceries at the store. You have the following bills: $100, $50, $20, $10, $5, $2, and $1. What are number of different sums of money that you can pull out of your if you pull out 3 bills without looking?

40. Combinations using your calculator “Math” p/b Scroll to “PRB” Select option “3” then hit “5” then hit “5” “10 choose 5” Clear your screen then enter “10” Now “enter”

41. Permutations or Combinations? 10. The number of 10 person committees formed from a group of 20 people. Your turn: (which is it?) 11. The number of ways 1 st, 2 nd, and 3 rd place trophies can be awarded to the top three contestants of 100 entrants. 12. The number of different 5 card “hands” that can be dealt from a pack of 52 cards. (in the game this “hand” is used in, you can rearrange your cards in your hand after they have been dealt). 13. The number of ways 700 people can line up while in the lunch line.

42. 14. 10 person committees that can be formed from a group of 20 people. Your turn: (calculate the number of different: 15. Ways to award 1 st, 2 nd, and 3 rd place trophies to the top three contestants of 10 entrants. 16. “Hands” that are possible when dealing a person 5 cards from a pack of 52 cards. (assume that in the game this “hand” is used in, you can rearrange your cards in your hand after they have been dealt). 17. The number of ways 20 people can line up while in the lunch line.

43. What if two of the letters are the same? Count the number of different 5-letter “words” that can be formed using the letters in the word “WAAAG”. “WAAAG”. AAAWG AAAWG AAAWG AAAWG AAAWG AAAWG To remove the “double counting” we must divide out the number of ways to permutate A, A and A divide out the number of ways to permutate A, A and A These are all examples of the same word and have been “double counted”. “double counted”. We must divide by 3! AAAWG AAAWG AAAWG AAAWG AAAWG AAAWG

44. Distinguishable Permutations We must “divide out” the permutations of the same object that result in indistinguishable arrangements. that result in indistinguishable arrangements. If a set 12 items to be permutated has 3 objects of one kind, and 4 objects of another kind, and 5 objects of another kind, and 4 objects of another kind, and 5 objects of another kind, then the number of distinguishable ways to arrange the 12 then the number of distinguishable ways to arrange the 12 items is: items is:

45. Distinguishable Permutations In general, we find the number of distinguishable permutations when using some elements that are indistinguishable when using some elements that are indistinguishable as follows: as follows: If a set N items to be permutated has A objects of one kind, and B objects of another kind, and C objects of another kind, and B objects of another kind, and C objects of another kind, and A + B + C = N then the number of distinguishable ways and A + B + C = N then the number of distinguishable ways to arrange the N items is: to arrange the N items is:

46. Your Turn: You have the following bills in your wallet: three $20’s, four $10’s, five $5’s, and six $1’s What is the number of distinct ways you could pay out the bills one at a time? 18.

47. Counting How many 5 card hands are there with all face cards (king, queen, jack). Which is it? This tells you the hands all have 5 face cards. So how many arrangements are there when taking 12 cards and many arrangements are there when taking 12 cards and picking 5 ? picking 5 ? Permutation: (different order  counted separately) Combination: (different order  not counted separately)

48. Your Turn: How many 5 card hands are there with no face cards? 19.

49. Counting Sometimes there are more than one condition that must be met. How many 5 card hands have all 5 cards the same suite (hearts, diamonds, spades, clubs). (hearts, diamonds, spades, clubs). Combination: (different order  not counted separately) 1 st we must pick the suite: 2 nd we must pick the 5 cards from that suite: By the multipication principle: total number hands is:

50. Counting How many 5 card hands have 2 aces ? Combination: (different order  not counted separately) 1 st we must pick the 2 aces: 2 nd we must pick the other 3 cards: (if the hand has exactly 2 aces, then we must not include the other two aces as possible picks) By the multipication principle: total number of hands is:

51. Your Turn: How many 5 card hands are there with two fives and two sixes? 20. Hint: (1) pick the 2 fives, (2) pick the 2 sixes, (3) Pick the last card. Use the multiplication rule.