#$&* course Mth 152 6/30 11:30 006. Cards
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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's. Self-critique: I got the solution, my calculator does not get to the total amount of the possible solutions however because the answer is too large. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:There are 4 possiblilities of choosing 5's, however you're only choosing two. That would then make it C(4,2). To then choose the 9's, we have 4 possibilities as well with only two total that we need to choose. Making that also C(4,2). To get the total amount of possibilities you must keep in mind the amount of cards that are not 9's nor 5's which would be 44 cards. To get the answer you would then again multiply the two combinations together as well as the remaining 44 cards. C(4,2)*(4,2)*44= 44 possibilities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique:OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:For this solution there are a total of four ways to choose the two 5's required cards needed. C(4,2). There are then four possible choices to choose from the 3 needed chosen cards. C(4,3). C(4,2)*(4,3)=4 possiblities with two 5's and three 9's. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3). Self-critique:OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For choosing two 5's, C(4, 2). You have two to pick out of the possible four. For the three facecards, there are C(4, 3). Since the way of finding the solution for a full house with three identical facecards and two 5's would be: C(4,2)*(4,3)*3= 12 possiblities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). Self-critique:OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the fullhouses, there is a possible C(4,2) and (4,3) for a pair of one number and three of a kind for another. You must then take into account there are a possible 13 pairs and 12 three of a kinds. This is based on the Fundamental Counting priciple, therefore, 13*12*C(4,2)*(4,3)= total amount of pairs and three of a kinds of full houses. The total answer would then be 624 possible full house hands. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses. Self-critique:I was not quite sure what you meant by denominations of cards until I had looked at the solution and saw the counting principle in order to figure out the problem. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In every deck of cards there are 13 possible cards for every suit there is. A flush is the same suit of cards. Since there are 13 cards per suit, you must choose 5 cards from each suit C(13, 5). With a total of 4 suits oer deck making the final solution: C(13,5)*4= confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique:The answer to this problem went on and on even with a calculator so I was not sure if the formula was enough or not. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:There are four cards each for the 5's, 6's, 7's, 8's, and 9's. Therefore from 5 to 9 there would be C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5= 1024 possible ways. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9. STUDENT QUESTION not sure I understand why is it not C(20,5) I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not… INSTRUCTOR RESPONSE There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations. However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts. That's not a straight, nor are most of the C(20, 5) combinations of these cards. C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).< Self-critique:OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:There are 10 denominations for a low card straight from ace to ten. There are 4^5 possibilities for each low denominations. Making the solution 10*4^5. Giving a total of 10,240 possible straights with ace as the high or low card. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one? INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card. Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination. ********************************************* Question: `q009. Using a standard deck, in how many ways is it possible to get a 5-card hand consisting of all face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are a total of three faces with four cards per facecard. Making a total of 12 facecards per deck. C(12, 5) 12! = / 35!