Problem set 8 (a) How many different hands are possible? Solution. There is one hand for each way of choosing 7 cards from a 52-card deck Therefore, the number of possible hands is by the subset rule (b)How many hands contain three pairs and no three-of-a-kind or four-of-a-kind? Solution. There is a bijection with sequences specifying The values of the pairs, which can be chosen in(3)ways The suits of the lowest-value pair, which can be chosen in(2)ways The suits of the middle-value pair, which can be chosen in(2)ways The suits of the highest-value pair, which can be chosen in ()ways The value of the remaining card, which can be chosen in 10 ways The suit of the remaining card, which can be chosen in 4 ways Thus, the number of seven-card poker hands containing three pairs and no three or four-of-a-kind is 13 By the Generalized Product rule (c) How many hands have all cards of the same suit? Solution. There is a bijection with sequences specifying: The suit of all the cards, which can be chosen in 4 ways The values of the cards, which can be chosen in(3)ways So the seven cards can be chosen in 4()ways (d) How many hands have 5 or more face cards?(The jacks, queens, and kings are the face cards) Solution. There is a bijection between hands with exactly k face cards and pairs consisting of: A set of k face cards, which can be chosen in(k)ways A set of 7-h numbered cards, which can be chosen in(-_k)ways� � � � � � � � � � � � � � � � � � Problem Set 8 3 (a) How many different hands are possible? Solution. There is one hand for each way of choosing 7 cards from a 52­card deck. Therefore, the number of possible hands is: 52 7 by the Subset Rule. (b) How many hands contain three pairs and no three­of­a­kind or four­of­a­kind? Solution. There is a bijection with sequences specifying: 13 • The values of the pairs, which can be chosen in ways. 3 • The suits of the lowest­value pair, which can be chosen in 4 2 ways. • The suits of the middle­value pair, which can be chosen in 4 2 ways. • The suits of the highest­value pair, which can be chosen in 4 2 ways. • The value of the remaining card, which can be chosen in 10 ways. • The suit of the remaining card, which can be chosen in 4 ways. Thus, the number of seven­card poker hands containing three pairs and no three or four­of­a­kind is: � � � �3 13 4 10 · 4 3 · 2 · By the Generalized Product Rule. (c) How many hands have all cards of the same suit? Solution. There is a bijection with sequences specifying: • The suit of all the cards, which can be chosen in 4 ways. 13 • The values of the cards, which can be chosen in ways. 7 13 So the seven cards can be chosen in 4 · ways. 7 (d) How many hands have 5 or more face cards? (The jacks, queens, and kings are the face cards.) Solution. There is a bijection between hands with exactly k face cards and pairs consisting of: • A set of k face cards, which can be chosen in 12 ways. k • A set of 7 − k numbered cards, which can be chosen in 40 ways. 7−k