challset2sol

# 3 c62 8 22 page 20115 we must classify the possible

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3! + C(6,2) 8! 2!2! ). page 201,15. We must classify the possible types of five letter “words”. type I. Four letters of one kind and another different letter. type II. Three letters of one kind and two of another. type III. Three letters of one kind and two other letters each of a different kind. type IV. Two letters of one kind, two letters of another kind and one other letter. type V. Two letters of one kind, and three different letters. Mississippi has one M, four I’s , four S’s and two P’s.

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5 For type I, there are C(2,1) x C(3,1) x 5 4 ! ! such “words”. For type II, there are C(2,1) x C(2,1) x 5 3 2 ! ! ! such “words”. For type III, there are C(2,1) x C(3,2) x 5 3 ! ! such “words”. For type IV, there are C(3,2) x C(2,1) x 5 2 2 ! ! ! such “words”. For type V, there are C(3,1) x 5 2 ! ! such “words”. Using the “or” principle, our answer will be the sum of all these cases. 8. page 189, 12 A standard 52 card deck has 13 types of cards, each type comes in four suits. a. The number of five-card poker hands with four aces, is C(4,4) x C(48,1). C(4,4) counts the number of ways to select four aces out of the four aces available. After the four aces are selected there are C(48,1) ways to select a non-ace. b. The number of five-card poker hands with four of a kind is C(13,1)xC(4,4)xC(48,1). From thirteen types of cards, pick one of them, i.e. C(13,1) ways, next pick four of the four cards of available of that type, i.e. C(4,4) ways, and with the forty-eight cards left pick one them, i.e. C(48,1) ways. c. The number of five-card poker hands with exactly two different pairs is C(13,2) x C(4,2) x C(4,2) x C(44,1). From thirteen types of cards, pick two of them, i.e. C(13,2) ways, next pick two of the four cards available of the first type, do the same with the second type, i.e. C(4,2) x C(4,2) ways to do both, and with the forty-four cards remaining select one more ,i.e. C(44,1) ways.
6 d. The number of five-card poker hands that are a full house is C(13,1)xC(4,3)xC(12,1)xC(4,2). From thirteen types of cards, pick one of them which will form the three of a kind, i.e. C(13,1) ways, next select three cards from the four available of this type. There are now twelve types of cards left, pick one of them to form the two of a kind, i.e. C(12,1) ways, next select two of the four cards available of this type. e. There are nine possible ways to get consecutive values. They are 2 3 4 5 6, 3 4 5 6 7, 4 5 6 7 8, 5 6 7 8 9, 6 7 8 9 10, 7 8 9 10 J, 8 9 10 J Q, 9 10 J Q K, 10 J Q K A. Each sequence can be obtained in C(4,1)xC(4,1)xC(4,1)xC(4,1)xC(4,1) ways, because each card comes in four suits. Using the “or” principle, we sum all of these up, and get 9 x 4 5 as our answer. Note that this answer includes straight flushes. f. The number of five-card poker hands that have no pairs, is C(13,5)xC(4,1)xC(4,1)xC(4,1)xC(4,1)xC(4,1) ways. From the thirteen types of cards available, choose five of them, i.e. C(13,5) ways. From each of the types you selected, choose one card from each.

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• Fall '06
• miller
• Numerical digit, ways

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