course Mth 152 ×jž‡²ŽÆæ“øw¸ã§Îä™ìÍŠùíÙ‚†˜Öôassignment #006
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20:13:08 `q001. Note that there are 8 questions in this assignment. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?
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RESPONSE --> In a deck of cards there are 4 fives. To be able to get 2 out of the 4 we would get C(4,2) If 4 of the deck are 5's then we would have 48 more cards to choose the other 3 cards from. C(4,2,) x (48,3) = confidence assessment: 2
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20:13:19 In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.
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RESPONSE --> o.k. self critique assessment: 3
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20:16:15 `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?
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RESPONSE --> using a standard deck of 52 cards again we have 4 fives and 4 nines, so we would have 44 cards left (4,2) x (4,2) x 44 confidence assessment: 3
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20:16:22 There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5. The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.
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RESPONSE --> o.k self critique assessment: 3
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20:19:02 `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?
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RESPONSE --> (4,2) x (9,3) = 6048 confidence assessment: 2
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20:19:13 There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).
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RESPONSE --> o.k. self critique assessment: 3
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20:22:06 `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?
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RESPONSE --> C(4,2) x C(4,3) x the three choices of face cards C(4,2) x C(4,3) x 3 confidence assessment: 3
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20:22:13 There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).
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RESPONSE --> o.k. self critique assessment: 3
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20:28:28 `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?
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RESPONSE --> there are four ways of getting a pair C(4,2) and four ways of getting three numbers that ard the same C(4,3) 52 cards in a deck divided by 4 ways to get pairs = 13 48 cards left divided by 4 ways to get the three numbers = 12 C(4,2) x C(4,3) x 13 x 12 confidence assessment: 2
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20:28:49 For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.
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RESPONSE --> o.k. self critique assessment: 3
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20:31:13 `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?
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RESPONSE --> 13 cards are in a suit C(13,5) there are four different suits in a deck of cards so htere would be C(13,5) x 4 confidence assessment: 3
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20:31:21 There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.
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RESPONSE --> o.k. self critique assessment: 3
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20:33:43 `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?
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RESPONSE --> There would be four each of 5,6,7,8, and 9 in a deck so 4x4x4x4x4 = 1024 confidence assessment: 3
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20:33:51 There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.
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RESPONSE --> o.k. self critique assessment: 3
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20:39:48 `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?
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RESPONSE --> A,1,2,3,4 1,2,3,4,5 2,3,4,5,6 3,4,5,6,7 4,5,6,7,8 = 10 x (4x4x4x4x4 ) 5,6,7,8,9 possible straights 6,7,8,9,10 7,8,9,10,J 8,9,10,J,Q 9,10,J,Q,K confidence assessment: 2
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20:39:55 There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.
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RESPONSE --> o.k. self critique assessment: 3
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