Chapter 9 Discrete Mathematics

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1 Section 9. Basic Combinatorics 355 Chapter 9 Discrete Mathematics Section 9. Basic Combinatorics Exploration. Six: ABC, ACB, BAC, BCA, CAB, CBA.. Approximately person out of 6, which would mean 0 people out of No. If they all looked the same, we would expect approximately 0 people to get the order right simply by chance. The fact that this did not happen leads us to reject the look-alike conclusion. 4. It is likely that the salesman rigged the test to mislead the office workers. He might have put the copy from the more expensive machine on high-quality bond paper to make it look more like an original, or he might have put a tiny ink smudge on the original to make it look like a copy. You can offer your own alternate scenarios. Quick Review Section 9. Exercises. There are three possibilities for who stands on the left, and then two remaining possibilities for who stands in the middle, and then one remaining possibility for who stands on the right: 3 # # 6.. Any of the four jobs could be ranked most important, and then any of the remaining three jobs could be ranked second, and so on: 4 # 3 # # Any of the five books could be placed on the left, and then any of the four remaining books could be placed next to it, and so on: 5 # 4 # 3 # # Any of the five dogs could be awarded st place, and then any of the remaining four dogs could be awarded nd place, and so on: 5 # 4 # 3 # # There are 3 # 4 possible pairs: K Q, K Q, K Q 3, K Q 4, K Q, K Q, K Q 3, K Q 4, K 3 Q, K 3 Q, K 3 Q 3, and K 3 Q There are 3 # 4 possible routes. In the tree diagram, B represents the first road from town A to town B, etc. 7. 9!36,880 (ALGORITHM) There are letters, where S and I each appear 4 times and P appears times. The number of distinguishable permutations is! 4!4!! 34, There are letters, where A appears 3 times and O and T each appear times. The number of distinguishable permutations is! 3!!!,663,00.. The number of ways to fill 3 distinguishable offices from a pool of 3 candidates is 3P 3 3! 0! 76.. The number of ways to select and prioritize 6 out of projects is P 6! 665,80. 6! 3. 4 # 3 # # # # # # A 6! 6 -! 6 # 5 # 4! 30 4! 9! 9 -! 9 # 8 # 7! 7 7! 0! 7!0-7! 0 # 9 # 8 # 7! 7! # 3 # # 0 0! 3!0-3! 0 # 9 # 8 # 7! 3! # 0 # 9 # 8 7! 3 # # 0 9. combinations 0. permutations. combinations. permutations (different roles) 3. There are 0 choices for the first character, 9 for the second, 6 for the third, then 5, then 8, then 7, then 6: 0 # 9 # 6 # 5 # 8 # 7 # 6 9,656, There are 36 choices for each character: ,466, There are 6 possibilities for the red die, and 6 for the green die: 6 # There are possibilities for each flip: B B B 3 C C C 3 C 4 C C C 3 C 4 C C C 3 C 4

2 356 Chapter 9 Discrete Mathematics C 3 5C 5 48C Choose 7 positions from the 0: 0! 0C 7 7!0-7! 0! 7!3! 77,50 3. Choose A and K, and cards from the other 50: C # 50 C # 50! 50 C!50 -! 50!!39! 37,353,738, C We either have 3,, or student(s) nominated: 6C 3 6 C 6 C We either have 3,, or appetizer(s) represented: 5C 3 5 C 5 C Each of the 5 dice have 6 possible outcomes: ! 36. 0C 8 8!0-8! 0! 5, 970 8!! *4*3* Since each topping can be included or left off, the total number of possibilities with n toppings is n. Since 048 is less than 4000 but 4096 is greater than 4000, Luigi offers at least toppings. 40. There are n subsets, of which n - are proper subsets ,765, True. 5! 3!5-3! 5! 3!! 300 5! 5!5-5! 5! 5!47!,598,960 48! 3!48-3! 48! 3!45! 7,96 8! 3!8-3! 8! 3!5! 56 a n a b a!n - a! a!b! n - b!b! a n b b 44. False. For example, a 5 is greater than a 5 b 0 4 b There are a 6 different combinations of vegetables. The total number of entreé-vegetable-dessert b 5 variations is 4 # 5 # The answer is D P 5 30,40. The answer is D. 47. np n The answer is B. n - n! 48. There are as many ways to vote as there are subsets of a set with 5 members. That is, there are 5 ways to fill out the ballot. The answer is C. 49. Answers will vary. Here are some possible answers: (a) Number of 3-card hands that can be dealt from a deck of 5 cards (b) Number of ways to choose 3 chocolates from a box of chocolates (c) Number of ways to choose a starting soccer team from a roster of 5 players (where position matters) (d) Number of 5-digit numbers that can be formed using only the digits and (e) Number of possible pizzas that can be ordered at a place that offers 3 different sizes and up to 0 different toppings. 50. Counting the number of ways to choose the eggs you are going to have for breakfast is equivalent to counting the number of ways to choose the ten eggs you are not going to have for breakfast. 5. (a) Twelve (b) Every 0 represents a factor of 0, or a factor of 5 multiplied by a factor of. In the product 50 # 49 # 48 #... # #, the factors 5, 0, 5, 0, 30, 35, 40, and 45 each contain 5 as a factor once, and 5 and 50 each contain 5 twice, for a total of twelve occurrences. Since there are 47 factors of to pair up with the twelve factors of 5, 0 is a factor of 50! twelve times. 5. (a) Each combination of the n vertices taken at a time determines a segment that is either an edge or a diagonal. There are n C such combinations. (b) Subtracting the n edges from the answers in (a), we find that nc - n!n -! - n nn - n - 3n - n. 53. In the nth week, 5 n copies of the letter are sent. In the last week of the year, that s 5 5.*0 36 copies of the letter. This exceeds the population of the world, which is about 6*0 9, so someone (several people, actually) has had to receive a second copy of the letter. 54. Six. No matter where the first person sits, there are the 3!6 ways to sit the others in different positions relative to the first person. 55. Three. This is equivalent to the round table problem (Exercise 4), except that the necklace can be turned upside-down. Thus, each different necklace accounts for two of the six different orderings. 56. The chart on the left is more reasonable. Each pair of actresses will require about the same amount of time to interview. If we make a chart showing n (the number of actresses) and n C (the number of pairings), we can see that chart allows approximately 3 minutes per pair throughout, while chart allows less and less time per pair as n gets larger. Number Number of Time per Pair Time per Pair n pairs n C Chart Chart

3 Section 9. The Binomial Theorem There are 5 C 3 635,03,559,600 distinct bridge hands. Every day has 60 # 60 # 4 86,400 seconds; a year has days, which is 3,556,736 seconds. Therefore it will 635,03,559,600 take about L 0,3 years. (Using 365 days 3,556,736 per year, the computation gives about 0,36 years.) 58. Each team can choose 5 players in C 5 79 ways, so there are 79 67,64 ways total. Section 9. The Binomial Theorem Exploration. 3C 0 3! 0!3!, 3C 3!!! 3, 3C 3! These are (in order) the!! 3, 3C 3 3! 3!0!. coefficients in the expansion of (ab) 3.. { }. These are (in order) the coefficients in the expansion of (ab) { }. These are (in order) the coefficients in the expansion of (ab) 5. Quick Review 9.. x xyy. a abb 3. 5x -0xyy 4. a -6ab9b 5. 9s st4t 6. 9p -4pq6q 7. u 3 3u v3uv v 3 8. b 3-3b c3bc -c x 3-36x y54xy -7y m 3 44m n08mn 7n 3 4. a 0 9 b x y 9 a 0 0 b x0 y 0 x 0 0x 9 y 45x 8 y 0x 7 y 3 0x 6 y 4 5x 5 y 5 0x 4 y 6 0x 3 y 7 45x y 8 0xy 9 y 0 5. Use the entries in row 3 as coefficients: (xy) 3 x 3 3x y3xy y 3 6. Use the entries in row 5 as coefficients: (xy) 5 x 5 5x 4 y0x 3 y 0x y 3 5xy 4 y 5 7. Use the entries in row 8 as coefficients: (pq) 8 p 8 8p 7 q8p 6 q 56p 5 q 3 70p 4 q 4 56p 3 q 5 8p q 6 8pq 7 q 8 8. Use the entries in row 9 as coefficients: (pq) 9 p 9 9p 8 q36p 7 q 84p 6 q 3 6p 5 q 4 6p 4 q 5 84p 3 q 6 36p q 7 9pq 8 q x y 0 a 0 0 b x0 y 0 a 0 b x9 y a 0 b x8 y a 9 b 9!!7! 9 # 8 # 36 a 5 b 5!!4! 5 # 4 # 3 # 4 # 3 # # 365 a b 66! 66!0! a 66 0 b 66! 0!66! a 0 3 b x7 y 3 a 0 4 b x6 y 4 a 0 5 b x5 y 5 a 0 6 b x4 y 6 a 0 7 b x3 y 7 a 0 8 b x y 8 Section 9. Exercises a b 4 a 4 0 b a4 b 0 a 4 b a3 b a 4 b a b a 4 3 b a b 3 a 4 4 b a0 b 4 a 4 4a 3 b 6a b 4ab 3 b 4 a b 6 a 6 0 b a6 b 0 a 6 b a5 b a 6 b a4 b a 6 3 a3 b 3 a 6 4 b a b 4 a 6 5 b a b 5 a 6 6 b a0 b 6 a 6 6a 5 b 5a 4 b 0a 3 b 3 5a b 4 6ab 5 b 6 3. x y 7 a 7 0 b x7 y 0 a 7 b x6 y a 7 b x5 y a 7 3 b x4 y 3 a 7 4 b x3 y 4 a 7 5 b x y 5 a 7 6 b x y 6 a 7 7 b x0 y 7 x 7 7x 6 y x 5 y 35x 4 y 3 35x 3 y 4 x y 5 7xy 6 y , , f(x)(x-) 5 x 5 5x 4 ( )0x 3 ( ) 0x ( ) 3 5x( ) 4 ( ) 5 x 5-0x 4 40x 3-80x 80x-3 8. g(x)(x3) 6 x 6 6x # 5 3 5x 4 # 3 5x # 3 4 6x # 0x # x 6 8x 5 35x 4 540x 3 5x 458x h(x)(x-) 7 (x) 7 7(x) 6 ( )(x) 5 ( ) 35(x) 4 ( ) 3 35(x) 3 ( ) 4 (x) ( ) 5 7(x)( ) 6 ( ) 7 8x 7-448x 6 67x 5-560x 4 80x 3-84x 4x-

4 358 Chapter 9 Discrete Mathematics 0. f(x)(3x4) 5 (3x) 5 5(3x) 4 # 4 03x 3 # 4 0(3x) # x # x 5 60x 4 430x x 3840x 04. (xy) 4 (x) 4 4(x) 3 y6(x) y 4(x)y 3 y 4 6x 4 3x 3 y4x y 8xy 3 y 4. (y-3x) 5 (y) 5 5(y) 4 ( 3x)0(y) 3 ( 3x) 0(y) ( 3x) 3 5(y)( 3x) 4 ( 3x) 5 3y 5-40y 4 x70y 3 x -080y x 3 80yx 4-43x x - y 6 x 6 6 x 5 - y 5 x 4 # - y 0 x 3 - y 3 5 x - y 4 6 x-y 5 - y 6 x 3-6x 5/ y / 5x y-0x 3/ y 3/ 5xy -6x / y 5/ y 3 x 3 4 x 4 4 x x # 3 4 x (x 3) 5 (x ) 5 5(x ) 4 x -0 5x -8 90x 6 70x 4 405x (a-b 3 ) 7 a 7 7a 6 ( b 3 )a 5 ( b 3 ) 35a 4 ( b 3 ) 3 35a 3 ( b 3 ) 4 a ( b 3 ) 5 7a( b 3 ) 6 ( b 3 ) 7 a 7-7a 6 b 3 a 5 b 6-35a 4 b 9 35a 3 b -a b 5 7ab 8 -b 7. Answers will vary. 8. Answers will vary. 9. a n b!n -! n n -!! n -!3n - n - 4! a n n - b 30. a n r b r!n - r! n - r!r! n - r!3n - n - r4! a n n - r b 3. n - r - n - r n -! r -!3n - - r - 4! n -! r!n - - r! rn -! rr -!n - r! n -!n - r r!n - rn - r -! rn -! n - rn -! r!n - r! r!n - r! r n - rn -! r!n - r! r!n - r! n r x 4x3x 8x 3x 9 # 3 0x 0x - # - 3 # x - # (a) Any pair (n, m) of nonnegative integers except for (, ) provides a counterexample. For example, n and m3: (3)!5!0, but!3!68. (b) Any pair (n, m) of nonnegative integers except for (0, 0) or any pair (, m) or (n, ) provides a counterexample. For example, n and m3: # 3! 6! 70, but,! # 3! # a n b a n b n n - n n - nn - n n n - n n n n! n -!3n - n - 4! n - n n n n 35. True. The signs of the coefficients are determined by the powers of the ( y) terms, which alternate between odd and even. 36. True. In fact, the sum of every row is a power of. 37. The fifth term of the expansion is n!!n -!!n -! n -!3n - n - 4! n! n -!! n -!! nn - n n a 8 The answer is C. 4 bx4 4 0x 4. n 38. The two smallest numbers in row 0 are and 0. The answer is B. 39. The sum of the coefficients of (3x-y) 0 is the same as the value of (3x-y) 0 when x and y. The answer is A. 40. The even-numbered terms in the two expressions are opposite-signed and cancel out, while the odd-numbered terms are identical and add together. The answer is D. 4. (a), 3, 6, 0, 5,, 8, 36, 45, 55. (b) They appear diagonally down the triangle, starting with either of the s in row. (c) Since n and n represent the sides of the given rectangle, then n(n) represents its area. The triangular number is / of the given area. Therefore, the nn triangular number is.

5 Section 9.3 Probability 359 (d) From part (c), we observe that the nth triangular nn number can be written as. We know that n () n n 0 n n n - n n - p n n n binomial coefficients are the values of n r for r0,,,3,...,n. We can show that nn n as follows: nn n nn -! n -! n!!n -! n!!n -! n So, to find the fourth triangular number, for example, 5 # 4 # 3! compute 4 5! 5!5 -!!3! 5 # (a). (Every other number appears at least twice.) (b) (c) No. (They all appear in order down the second diagonal.) (d) 0. (See Exercise 44 for a proof.) (e) All are divisible by p. (f) Rows that are powers of :, 4, 8, 6, etc. (g) Rows that are less than a power of :, 3, 7, 5, etc. (h) For any prime numbered row, or row where the first element is a prime number, all the numbers in that row (excluding the s) are divisible by the prime. For example, in the seventh row ( ) 7,, and 35 are all divisible by The sum of the entries in the nth row equals the sum of the coefficients in the expansion of (xy) n. But this sum, in turn, is equal to the value of (xy) n when x and y: n n a n 0 b n 0 a n b n- a n b n n p a n n b 0 n a n 0 b a n b a n b p a n n b n 0 n n n - - p n ( ) n n n n - - n 0 - n n p - n n n Section 9.3 Probability Exploration.. 3. n 0 n 4 n p n n n p p p p p p P(±) Pantibody present and 5. P(antibody present ±) P (A little more than chance in 4.) L 0.86 Quick Review C 5,598, C !0 present absent present absent present absent p p p 0.05 p p p p 0.05 p p p p p

6 360 Chapter 9 Discrete Mathematics 8. 5 P C 3 0C 3 5C 0C 5! 3!! 0! 3!7! 5!!3! 0!!8! 9 Section 9.3 Exercises For # 8, consider ordered pairs (a, b) where a is the value of the red die and b is the value of the green die. 4. E{(3, 6), (4, 5), (5, 4), (6, 3)}: P(E) E{both dice even}, {both dice odd}: 3 # 3 3 # 3 P(E) E{(, ), (3, ), (3, ), (4, ), (4, ), (4, 3), (5, ), (5, ), (5, 3), (5, 4), (6, ), (6, ), (6, 3), (6, 4), (6, 5)}: 5 P(E) E{(, ), (, ),...,(6, ), (6, 3)} 30 P(E) # 3 5. P(E) # 3 6. P(E) E{(, ), (, ), (, 4), (, 6), (, ), (, 3), (, 5), (3, ), (3, 4), (4, ), (4, 3), (5, ), (5, 6), (6, ), (6, 5)} 5 P(E) E{(, 6), (, 5), (3, 4), (4, 3), (5, ), (5, 6), (6, ), (6, 5)}: 8 P(E) (a) No The numbers do not add up to. (b) There is a problem with Alrik s reasoning. Since the gerbil must always be in exactly one of the four rooms, the proportions must add up to, just like a probability function. 0. Since 430, we can divide each number in the ratio by 0 and get the proportions relative to the whole. The table then becomes Compartment A B C D Proportion Yes, this is a valid probability function.. P(B or T)P(B)P(T) P(R or G or O)P(R)P(G)P(O) P(R)0. 4. P(not R)-P(R) P[not (O or Y)]-P(O or Y) -(0.0.) P[not (B or T)]-P(B or T)-(0.30.) P(B and B )P(B ) # P(B )(0.3)(0.3) P(O and O )P(O ) # P(O )(0.)(0.) P[(R and G ) or (G and R )]P(R ) # P(G ) P(G ) # P(R )(0.)(0.)(0.)(0.) P(B and Y )P(B ) # P(Y )(0.3)(0.)0.06. P(neither is yellow)p(not Y and not Y ) P(not Y )# P(not Y )(0.8)(0.8)0.64. P(not R and not O )P(not R )# P(not O ) (0.8)(0.9) There are 4 C 6 34,596 possible hands; of these, only consists of all spades, so the probability is. 34, Of the 4 C 6 34,596 possible hands, one consists of all spades, one consists of all clubs, one consists of all hearts, and one consists of all diamonds, so the probability is 4 34,596 33, Of the 4 C 6 34,596 possible hands, there are 4C 4 # 0 C 90 hands with all the aces, so the probability 90 is 34, There are C # C ways to get both black jacks and 4 other cards. Similarly, there are 735 ways to get both red jacks. These two numbers together count twice the C # C # 0 C 90 ways to get all four jacks. Therefore, altogether we have # ,440 distinct ways to have both bowers, so the probability is 4,440 34, (a) A (b) 0.3 (c) 0. (d) 0. PA and B (e) Yes. P(A B) PB 8. (a) A B B PA (b) 0.5 (c) 0. (d) 0. PA and B (e) No. P(A B) Z PA PB P(John will practice)(0.6)(0.8)(0.4)(0.4)0.64

7 Section 9.3 Probability (a) P(meatloaf is served) (b) P(meatloaf and peas are served)(0.0)(0.70) 0.4 (c) P(peas are served)(0.0)(0.70)(0.80)(0.30) If all precalculus students were put on a single list and a name then randomly chosen, the probability P(from Mr. Abel s class girl) would be 6. But when one of the two classes is selected at random, and then a student from this class is selected, P(from Mr. Abel s class girl) Pgirl from Mr. Abel s class Pgirl 3. Within each box, any of the coins is equally likely to be chosen and either side is equally likely to be shown. But a head in the -coin box is more likely to be displayed than a head in the 3-coin box. PH from -coin box P(from -coin box H) PH 3 5 a ba 0 b a ba 0 b a ba0 5 b a ba3 4 b a ba3 4 b a ba3 6 b 3 5 0C C P(none are defective) 4 C 0 # (0.037) 0 (0.963) 4 (0.963) (a) P(cardiovascular disease or cancer) (b) P(other cause of death) P(Yahtzee) 6 # a 5 6 b The sum of the probabilities is greater than an impossibility, since the events are mutually exclusive. 38. P(at least one false positive) -P(no false positives)-(0.993) (a) P(a woman) (b) P(went to graduate school) (c) P(a woman who went to graduate school) (a) (b) (c) 4C C 8 5, C 5 # 6 C 3 40,040 0C 8 5, C 6 # 6 C 4 C 7 # 6 C 4 C 8 # 6 C 0 0C 8 45,045 0, , C This cannot be true. Let A be the event that it is cloudy all day, B be the event that there is at least hour of sunshine, and C be the event that there is some sunshine, but less than hour. Then A, B, and C are mutually exclusive events, so P(A or B or C)P(A)P(B)P(C) P(C)P(C). Then it must be the case that P(C)0. This is absurd; there must be some probability of having more than 0 but less than hour of sunshine. 43. P(HTTTTTTTTT) 44. P(THTTTTHTTT) 45. P(HHHHHHHHHH) a b a b 0 04 a b P(9 H and T) 0C # a 0 b P( H and 8 T) 0C # a 0 b ,640 5, P(3 H and 7 T) 0C 3 # a 0 b P(at least one H)-P(0 H) - 0 C 0 # a 0 b P(at least two H)-P(0 H)-P( H) - 0 C 0 # a 0 b - 0 C # a b False. A sample space consists of outcomes, which are not necessarily equally likely. 5. False. All probabilities are between 0 and, inclusive. 53. Of the 36 different, equally-likely ways the dice can land, 4 ways have a total of 5. So the probability is 4/36/9. The answer is D. 54. A probability must always be between 0 and, inclusive. The answer is E. 55. P(B and A)P(B) P(A B), and for independent events, P(B and A)P(B)P(A). It follows that the answer is A. 56. A specific sequence of one heads and two tails has probability (/) 3 /8. There are three such sequences. The answer is C.

8 36 Chapter 9 Discrete Mathematics 57. (a) (b) (0.37)(0.37)(0.37) 0.05 (c) No. They are more apt to share bagel preferences if they arrive at the store together. 58. (a) P(at least one king)-p(no kings) 48C 5-8,47 5C 5 54,45 L 0.34 (b) The number of ways to choose, e.g., 3 fives and jacks is 4 C 3 # 4 C. There are 3 P 3 # different combinations of cards that can make up the full house, so 3 # # 4 C 3 # 4 C P(full house) 6 5C L (a) We expect (8)(0.3).84 (about ) to be married. (b) Yes, this would be an unusual sample. (c) P(5 or more are married) 8 C 5 # (0.3) 5 (0.77) 3 8 C 6 # (0.3) 6 (0.77) 8 C 7 # (0.3) 7 (0.77) 8 C 8 # (0.3) 8 (0.77) %. 60. (a) 3 C 7 # (0.75) 7 (0.5) % (b) a 3C k # 0.75 k k L % (c) The university s graduation rate seems to be exaggerated; at least, this particular class did not fare as well as the university claims. 6. (a) $.50 (b) 0 6. (a) 9,366,89 L (b) Value Probability (c) 7 k 0 Type of Bagel 3 # 6 - # ,366,809 9,366,89 3,999, ,366,89 3,999,990 # 0 9,366,89-0 # 0 a - 9,366,89 b L (d) In the long run, Gladys is losing $5.73 every time she buys the ten tickets. Given the low probability of a positive payoff, she stands to lose a lot of money if she does this often. Section 9.4 Sequences Probability Plain 0.37 Onion 0. Rye 0. Cinnamon Raisin 0.5 Sourdough 0.5 Quick Review (5-) [(6(5-)4] () # a a 0 5(0-)33 7. a 0 5 # a 0 a ba b a 4 3 ba 5 b a a Section 9.4 Exercises For # 4, substitute n, n,..., n6, and n00.., 3, 4 3, 5 4, 6 5, 7 6 ; ,, 4 5, 3, 4 7, ; , 6, 4, 60, 0, 0; 999, , 6, 6, 4, 0, 6; 9500 For #5 0, use previously computed values of the sequence to find the next term in the sequence. 5. 8, 4, 0, 4; , 7, 7, 7; 67 7., 6, 8, 54; ,.5, 3, 6; 96 9.,,, 0; 3 0., 3,, 4; 3. lim so the sequence diverges. nsq n q,. lim so the sequence converges to 0. nsq n 0, 3., 4, 9, 6,..., n,... lim so the sequence converges to 0. nsq n 0, 4. lim 3n - q, so the sequence diverges. nsq 5. Since the degree of the numerator is the same as the degree of the denominator, the limit is the ratio of the 3n - leading coefficients. Thus lim The nsq - 3n -. sequence converges to. 6. Since the degree of the numerator is the same as the degree of the denominator, the limit is the ratio of the n - leading coefficients. Thus lim The sequence nsq n. converges to. 7. lim so the sequence converges nsq 0.5n lim a n nsq b 0, to lim so the sequence diverges. nsq.5n lim a 3 n nsq b q,

4/17/2012. NE ( ) # of ways an event can happen NS ( ) # of events in the sample space

4/17/2012. NE ( ) # of ways an event can happen NS ( ) # of events in the sample space I. Vocabulary: A. Outcomes: the things that can happen in a probability experiment B. Sample Space (S): all possible outcomes C. Event (E): one outcome D. Probability of an Event (P(E)): the likelihood