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Chapter 11

Chapter 11. Counting Methods. Section 11.1. Systematic Listing. Counting Methods. Create a list systematically Create a table Can only be used in two-part tasks Create a tree Used in tasks involving three or more parts Can be tedious if task involves many parts Other techniques

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Chapter 11

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  1. Chapter 11 Counting Methods

  2. Section 11.1 Systematic Listing

  3. Counting Methods • Create a list systematically • Create a table • Can only be used in two-part tasks • Create a tree • Used in tasks involving three or more parts • Can be tedious if task involves many parts • Other techniques • Fundamental Counting Principle (FCP) • Permutations • Combinations Section 11.1

  4. Your Plan for Counting • How many parts does the task have? • Are there any restrictions? • Is repetition allowed? • If repetition is not allowed, does the problem indicate that {a,b,c} is different from {b,a,c}? • If not, compute using combinations, FCP, or other methods • If so, compute using permutations, FCP, or other methods • If repetition is allowed, then use FCP, lists, trees, or other methods Section 11.1

  5. Systematic Listing Method Jo packed 2 shirts (red, white) and 3 slacks (gray, black, khaki) for her trip to NYC • How many different outfits are possible? • Red shirt with gray slacks {R,G} • Red shirt with black slacks {R,B} • Red shirt with khaki slacks {R,K} • White shirt with gray slacks {W,G} • White shirt with black slacks {W,B} • White shirt with khaki slacks {W,K} Section 11.1

  6. Table Method Jo packed 2 shirts (red,white) & 3 slacks (gray,black,khaki) Slacks Shirts Again, we have 6 outfits Section 11.1

  7. Let’s add more to Jo’s bag . . . Let’s say Jo packed 2 shirts (red, white), 3 slacks (gray, black, khaki), and 2 shoes (dress, tennis) How many different outfits are possible? 12 possible outfits Section 11.1

  8. Gray Red Black Khaki Gray White Black Khaki Tree Method ShirtsSlacksShoes 3 parts Dress Tennis Dress Tennis Dress Tennis Dress Tennis Dress Tennis Dress Tennis • {R,G,D} • {R,G,T} • {R,B,D} • {R,B,T} • {R,K,D} • {R,K,T} • {W,G,D} • {W,G,T} • {W,B,D} • {W,B,T} • {W,K,D} • {W,K,T} Section 11.1

  9. How Many Outcomes Are Possible When Tossing 3 Coins? H H T H H T T H H T H T T T 3 parts 1st coin2nd coin3rd coin • HHH • HHT • HTH • HTT • THH • THT • TTH • TTT It does not matter if the coins are alike or different! Section 11.1

  10. Selection Process The NVCC Math Club has five members: Andy, Bill, Cathy, David, Ellen In how many ways can the club select: • a president? 5 ways (any one of the members) • a president and a secretary if no one may hold more than one position? 20 ways Note that (Cathy, Andy) means Cathy is President and Andy is Secretary, which is NOT the same as (Andy, Cathy) Here the order IS important so we use parentheses instead of brackets Section 11.1

  11. Selection Process (continued) In how many ways can the math club select a president and a secretary if no one may hold more than one position and the secretary must be a man? 12 ways Always take care of restrictions first! Section 11.1

  12. Selection Process (continued) In how many ways can the math club select a president, a secretary and a treasurer if no one may hold more than one position, the secretary must be a woman and the treasurer must be a man? Bill David Bill David Andy David Andy David Andy Bill David Andy Bill Andy Bill Andy Bill David Cathy Ellen Cathy Ellen Ellen Cathy Ellen Cathy Andy Bill Cathy David Ellen 18 ways Section 11.1

  13. How Many Outcomes Are Possible When Rolling a Pair of Dice? • How many parts to the task? • 2 • How many outcomes for each part? • 6 for part 1 (die 1) • 6 for part 2 (die 2) • Restrictions? • No • Is repetition allowed? • Yes • That is, the outcome on die 1 can match the outcome on die 2 Section 11.1

  14. How Many Outcomes Are Possible When Rolling a Pair of Dice? Blue Die Note that (2,4) is NOT the same as (4,2) Here, the order IS important, so we use parentheses instead of brackets Red Die Section 11.1 36 outcomes

  15. Section 11.2 The Fundamental Counting Principle

  16. Uniformity Criterion A multiple-part task is said to satisfy the uniformity criterion if the number of choices for any particular part is the same no matter which choices were selected for previous parts Uniform or Non-uniform? • Michelle’s printer allows optional settings with a panel of 4 on-off switches in a row. How many different settings can she select if no 2 adjacent switches can both be off? • How many possible non-repeating 3-digit numbers exist with digits from the set {1,2,3}? Section 11.2

  17. 1 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 1 1 Michelle’s Options 4 parts 1st switch2nd switch3rd switch 4th switch • 0101 • 0110 • 0111 • 1010 • 1011 • 1101 • 1110 • 1111 0=off 1=on Not uniform: The 2nd switch can be done in 1 way if the 1st switch is off, but it can be done in 2 ways when the 1st switch is on Section 11.2

  18. 2 3 1 3 2 1 3 2 3 1 1 2 3 2 1 Non-repeating 3-digit numbers 3 parts 1st digit2nd digit3rd digit • 123 • 132 • 213 • 231 • 312 • 321 UNIFORM!Each part has the same number of options no matter what number was chosen on the previous part Section 11.2

  19. Fundamental Counting Principle When a task consists of k separate parts and satisfies the uniformity criterion, if • the first part can be done in n1ways, • the second part can be done in n2 ways, • the third part can be done in n3 ways, • …, and • the kth part can be done in nk ways, then the total number of ways to complete the task is given by the product n1 •n2 •n3 •…•nk Translate “and” as multiplication Translate “or” as addition Section 11.2

  20. FCP Method – Outfits Jo packed 2 shirts (red,white) and 3 slacks (gray,black,khaki) • How many different outfits are possible? 2 ● 3 = 6 shirt & slacks Let’s say Jo packed 2 shirts (red, white), 3 slacks (gray, black, khaki), and 2 shoes (dress, tennis) • How many different outfits are possible? 2  3  2 = 12 shirt & slacks & shoes Section 11.2

  21. FCP Method – NVCC Math Club In how many ways can the club select: • a president and a secretary if no one may hold more than one position? 5 • 4 = 20 ways Pres & Sec • a president and a secretary if no one may hold more than one position and the secretary must be a man? 3 • 4 = 12 ways Sec & Pres • a president,a secretary and a treasurer if no one may hold more than one position, the secretary must be a woman and the treasurer must be a man? 3 • 2 • 3 = 18 ways Treas & Sec & Pres • a president,a secretary and a treasurer if no one may hold more than one position? 5 • 4 • 3 = 60 ways Pres & Sec & Treas Section 11.2

  22. FCP – “Combination” Locks How many distinct “combinations” can be created for a briefcase lock with three tumblers that use the digits 0 through 9 for the following: • Each digit can be repeated • 10  10  10 = 1,000 combinations • 1st 2nd 3rd • Each digit cannot be repeated • Once a digit is used, then one less digit available for the subsequent tumblers, so 10  9  8 = 720 combinations • 1st 2nd 3rd • Two consecutive tumblers cannot have the same digit The digit used on the first tumbler cannot be used for the second tumbler but can be repeated for the third tumbler, so • 10  9  9 = 810 combinations • 1st 2nd 3rd Section 11.2

  23. FCP – Drawing Cards If two cards are drawn from a standard 52-card deck without replacement, how many ways is it possible to obtain the following: • A black card on the first draw and a diamond on the second draw • These are disjoint sets, so we have a two-part task • 26  13 = 338 ways 1st 2nd black AND ♦ • A heart on the first draw and an ace on the second draw Break up into two disjoint events (the first card could be A) • 1  3 + 12  4 = 3 + 48 = 51 ways 1st 2nd 1st 2ndA♥ AND another A OR another♥ AND any A Section 11.2

  24. 3! = 3 • 2 • 1 = 6 6! = 6 • 5 • 4 • 3 • 2 • 1 = 720 6! – 3! =720 – 6 = 714 (6 – 4)! =2! = 2 • 1 = 2 0! = 1 Factorial Practice n! = n(n – 1)(n – 2)… • 2 • 1 Section 11.2

  25. FCP or Factorials The Math Club members are taking a trip to a conference and get five consecutive seats on the airplane. How many seating arrangements can be made? Fundamental Counting Principle method Factorial notation 5 • 4 • 3 • 2 • 1 = 5! = 240 seating arrangements Section 11.2

  26. FCP Does Not Apply How many seating arrangements can be made for the five club members given the following restrictions: • Andy and Ellen must sit together • Cathy and David must not sit together • Bill or David must be in the first seat 2nd seat 3rd seat 4th seat 5th seat David Cathy Ellen Andy Ellen Andy Cathy David Bill Cathy Ellen Andy Ellen Andy Bill Cathy Cathy X David X David  David  Cathy  Cathy  David X Cathy X Cathy  Bill  Cathy  Ellen  Andy  Cathy  Cathy  Bill  Andy Cathy David Ellen Andy Bill Ellen Ellen Andy Ellen Andy Ellen Andy Ellen Andy Cathy Ellen Andy 1st seat Bill David 12 ways Section 11.2

  27. Section 11.3 Permutations and Combinations

  28. Permutations and Combinations • Permutation • nPr or P(n,r) • The number of arrangements of n different things taken r at a time where r  n • ABC and BCA are different arrangements • Order IS important • No repetition allowed • Cue words – arrangement, schedule, pick in a particular order • Combination • nCr or C(n,r) • The number of subsets of n different things taken r at a time where r  n • ABC and BCA are NOT different subsets • Order NOT important • No repetition allowed • Cue words – set, subset, group, sample, selection, choose Section 11.3

  29. Permutation Combination Formulas OR OR Section 11.3

  30. Regular Formula = 5(5 – 1)(5 – 3 +1) = 5 • 4 • 3 = 60 Factorial Formula Evaluate 5P3 = P(5,3) Section 11.3

  31. Regular Formula Factorial Formula Evaluate 5C3 = C(5,3) We divide by r! to eliminate the repetition Section 11.3

  32. NVCC Math Club In how many ways can the club select a president, a secretary and a treasurer if no one may hold more than one position? Here, the order IS important  (Pres, Sec, Treas) Andy is President Bill is President Cathy is President David is President Ellen is President (A,B,C) (A,B,D) (A,B,E) (A,C,D) (A,C,E) (A,D,E) (A,C,B) (A,D,B) (A,E,B) (A,D,C) (A,E,C) (A,E,D) (B,A,C) (B,A,D) (B,A,E) (B,C,D) (B,C,E) (B,D,E) (B,C,A) (B,D,A) (B,E,A) (B,D,C) (B,E,C) (B,E,D) (C,A,B) (C,A,D) (C,A,E) (C,B,D) (C,B,E) (C,D,E) (C,B,A) (C,D,A) (C,E,A) (C,D,B) (C,E,B) (C,E,D) (D,A,B) (D,A,C) (D,A,E) (D,B,C) (D,B,E) (D,C,E) (D,B,A) (D,C,A) (D,E,A) (D,C,B) (D,E,B) (D,E,C) (E,A,B) (E,A,C) (E,A,D) (E,B,C) (E,B,E) (E,C,D) (E,B,A) (E,C,A) (E,D,A) (E,C,B) (E,D,B) (E,D,C) 5P3 = 60 ways Section 11.3

  33. {A,B,C} {A,B,D} {A,B,E} {A,C,D} {A,C,E} {A,D,E} {A,C,B} {A,D,B} {A,E,B} {A,D,C} {A,E,C} {A,E,D} {B,A,C} {B,A,D} {B,A,E} {B,C,D} {B,C,E} {B,D,E} {B,C,A} {B,D,A} {B,E,A} {B,D,C} {B,E,C} {B,E,D} {C,A,B} {C,A,D} {C,A,E} {C,B,D} {C,B,E} {C,D,E} {C,B,A} {C,D,A} {C,E,A} {C,D,B} {C,E,B} {C,E,D} {D,A,B} {D,A,C} {D,A,E} {D,B,C} {D,B,E} {D,C,E} {D,B,A} {D,C,A} {D,E,A} {D,C,B} {D,E,B} {D,E,C} {E,A,B} {E,A,C} {E,A,D} {E,B,C} {E,B,E} {E,C,D} {E,B,A} {E,C,A} {E,D,A} {E,C,B} {E,D,B} {E,D,C} NVCC Math Club In how many ways can the club choose three members to form a committee? The order in which the members are chosen doesn’t matter, so ABC = ACB = BAC = BCA = CAB = CBA for all possible combinations, so we divide out the repetition as indicated by the combination formula 5C3 = P(5,3) = 60 = 10 ways 3! 6 Section 11.3

  34. Five-Card Poker How many five-card poker hands can be created from a standard 52-card deck? The order in which the cards are dealt is NOT important, so you are choosing a 5-card subset Therefore, this is a combination problem P(52,5) = 52 • 51 • 50 • 49 • 48 = 311,875,200 5! 5  4  3  2  1 120 = 2,598,960 five-card poker hands 52C5= Section 11.3

  35. Five-Card Poker (continued) How many five-card poker hands can be created from a standard 52-card deck that have three red cards and two black cards? • This is a two-part task  look at each task and multiply Task 1 Choose 3 red out of 26 in the deck Task 2 Choose 2 black out of 26 in the deck AND n(3 red) = 26C3 = 26! = 26 • 25 •24 3! 23! 3 • 2 • 1 = 2600 n(2 black) = 26C2 = 26! = 26 • 25 2! 24! 2 • 1 = 325 2600 • 325 = 845,000 hands with 3 red & 2 black cards Section 11.3

  36. Five-Card Poker (continued) How many five-card poker hands can be created from a standard 52-card deck that have one heart, two clubs and two diamonds? • This is a three-part task  look at each task and multiply Task 1 Choose 1  out of 13 in the deck Task 2 Choose 2 ♣ out of 13 in the deck Task 3 Choose 2 ♦ out of 13 in the deck AND AND n(1) = 13C1 = 13! 1! 12! = 13 n(2♣) = 13C2 = 13! = 13 • 12 2! 11! 2 • 1 = 78 n(2♦) = 13C2 = 13! = 13 • 12 2! 11! 2 • 1 = 78 So 13 • 78 • 78 = 79,092 hands with 1, 2♣, 2♦ cards Section 11.3

  37. Section 11.5 Counting Problems Involving “Not” and “Or”

  38. Recall from Set Theory . . . • Complements Principle of Counting: If A is any set within the universal set U then n(A) = n(U) – n(A') • Special Additive Counting Principle: If A and B are disjoint sets, then n(AB) = n(A) + n(B) • General Additive Counting Principle: If A and B are any two sets, disjoint or not, then n(AB) = n(A) + n(B) – n(AB) Remember that union corresponds to the word “or” whereas intersection corresponds to the word “and” Section 11.5

  39. Complements Principle How many five-card poker hands can be formed from a standard 52-card deck which contain at least one heart?  or  or  or  or  This involves five calculations Instead, let’s find the complement and subtract from the total number of possible five-card poker hands that is, n(at least 1) = n(five-card hands) – n(0) To find the complement, no hearts, we want to choose five cards from the 39 cards in the deck that are ♠♣♦ So n(at least 1) = 52C5 – 39C5 = 52 • 51 • 50 • 49 • 48 _ 39  38  37  36  355  4  3  2  1 5  4  3  2  1 = 2,598,960 – 575,757 = 2,023,203 five-card hands with at least 1 Section 11.5

  40. Special Additive Principle How many five-card poker hands can be formed from a standard 52-card deck which contain either all hearts or all black cards? Since hearts cannot be black, we have two disjoint sets So we want to find n() + n(five black) 13 s in the deck, choose 5 13 • 12 • 11• 10• 9 5  4  3  2  1 n() = 13C5 = = 1,287 26 black cards in the deck, choose 5 26 • 25 • 24• 23• 22 5  4  3  2  1 n(five black) = 26C5 = = 65,780 Thus n() + n(five black) = 67,067 hands Section 11.5

  41. General Additive Principle How many five-card poker hands can be formed from a standard 52-card deck which contain either all red or all face cards? Since face cards can also be red, we do NOT have disjoint sets So we want to find n(five red) + n(five face) – n(five red face) 26 red cards in the deck, choose 5 26 • 25 • 24• 23• 22 5  4  3  2  1 n(five red) = 26C5 = = 65,780 12 face cards in the deck, choose 5 12 • 11 • 10• 9• 8 5  4  3  2  1 = 792 n(five face) = 12C5 = 6 red face cards in the deck, choose 5 6 • 5 • 4• 3• 2 5  4  3  2  1 = 6 n(five red face) = 6C5 = Section 11.5 So n(five red) + n(five face) – n(five red face) = 66,566 hands

  42. General Additive Principle How many ways can a red die and a blue die be rolled so that either the sum of the dice is at least 8 or the red die shows a 5? 15 bold/shaded table entries have a sum of at least 8 6 red/circled table entries are those showing a red 5 4 bold/shaded, red/circled table entries are those with a sum > 8 AND show a red 5 Sum>8 + Red5 – Both = 15 + 6 – 4 Blue Die Red Die 17 outcomes Section 11.5

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