#$&* course MTH 152 2/20 7 006. Cards*********************************************
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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can use the same sort of method to solve this problem as we did the previous problem. If we are trying to create a hand with two 5's we still have an equation of C(4,2), and if we are trying to create a hand with two 9's we have the same exact equation of C(4,2) yet again. This is because we are trying to get the same number of 9's as we are 5's, and because they have the same number of cards in the deck. Because we are creating a five card hand and two 5's and two 9's only leave us with one remaining card to choose, we have 44 remaining cards that are not 5's or 9's to choose from. This gives us an equation of C(4,2) * C(4,2) * 44. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* 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: In order to get two 5's from a standard deck of cards, we still must use the same C(4,2) equation of the previous two problems. In order to get three 9's however, we must account for the third 9 that we are going to choose from the deck. This time, we have an equation of C(4,3) for our 9 cards instead of C(4,2). Our final, overall equation is C(4,2) * C(4*3), which will gives us the number of ways it is possible to get a “full house” consisting of two 5's and three 9's. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3). Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can solve this problem in the same exact way that we solved the previous problem. C(4,2) is the equation with which we can choose two 5 cards out of a standard deck of cards, and C(4,3) is the equation that will work when we are trying to find three of any of the face cards in a deck. Our equation is C(4,2) * C(4,3), and accounting for the three choices of the face cards we are choosing, our final equation is C(4,2) * C(4,3) * 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: There are C(4,2) ways that we can choose two cards of any suit or “denomination” and C(4,3) ways that we can choose three cards of any suit or “denomination.” There are 13 possible choices for the initial pair of cards chosen, and 12 possible choices left for the three-of-a-kind when it is chosen. This gives us a final equation of 13 * 12 * C(4,2) * C(4,3). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 13 cards in any suit of a standard deck of cards, and 4 suits that make up an entire deck. In order to choose 5 cards of the same suit, we must choose 5 out of any of 13 cards of any of the 4 suits. This gives us an equation of C(13,5) * 4. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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 4 suits in any deck of cards, and in order to get a 5-card straight consisting of exactly one each of the denominations 5, 6, 7, 8, and 9, we must account for both of those factors. There are 4 each of all 5 cards we are trying to obtain, so we have an equation of 4^5, which is equal to 1,024 ways in which it is possible to get a 'straight' consisting of exactly one of each of the denominations 5, 6, 7, 8, and 9. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9. STUDENT QUESTION not sure I understand why is it not C(20,5) I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not… INSTRUCTOR RESPONSE There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations. However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts. That's not a straight, nor are most of the C(20, 5) combinations of these cards. C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).< Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 10 possible cards from each suit of the deck that can be considered low cards and that can be used as the beginning to any straight. As we discovered in the previous problem, there are 4^5 possible straights to be had for each low card of the deck. This gives us an equation of 10 * 4^5, or 10,240 ways in which it is possible to get a 'straight' consisting of five cards of consecutive denominations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. " Self-critique (if necessary): ------------------------------------------------ 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 possible cards from each suit of the deck that can be considered low cards and that can be used as the beginning to any straight. As we discovered in the previous problem, there are 4^5 possible straights to be had for each low card of the deck. This gives us an equation of 10 * 4^5, or 10,240 ways in which it is possible to get a 'straight' consisting of five cards of consecutive denominations. 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!