#$&* course Mth 152 I'm having problems with computing problems such as C (4,2) * C (4,2) * 44. Can you help me by explaining how to go through it step by step or tell me where I can find it in the book. I can't find it in the book??? 006. Cards
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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C (4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C (4,2) * C (48, 3) ways to obtain exactly two 5's. Self-critique: 2 - I am not sure if I am getting the right answer for these??? I’m going to continue working them but If I am not, I will redo them if you let me know what I am doing wrong. ------------------------------------------------ Self-critique rating: 1
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Given Solution: There are C (4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards, which are neither 9 nor 5. The total number of possible ways is therefore C (4, 2) * C (4, 2) * 44. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are looking for a full house hand that consists of two 5’s and three 9’s. C (4,2) * C (4,3) 8 * 3 24 ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* 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: C (4, 2) * C (4, 3) * 3 72 ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3). Self-critique: okay ------------------------------------------------ Self-critique rating: okay ********************************************* Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are looking for a full house of 5 cards that consist of at least two cards of one suit and three of another. Using the counting principle to finds the specific suits we use: C (4,2) * C (4,3). In the deck there are 13 possible kinds of cards that could be used in either the two pair of the three pair. 13 possibilities for the first pair means there are only 12 possibilities for the next pair and so on to 11 possibilities if we were making a third pair of cards. Therefore we want to multiply C (4,2) * C (4,3) by the possibilities of 12 and 13. C (4,2) * C (4,3) * 13 * 12 3744 choices of full houses confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: okay ------------------------------------------------ Self-critique rating: okay ********************************************* 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 each suit. If we want to get a flush of five cards from one suit there will be 8 cards left in that suit. To find the possibilities of the 5 cards of the 13 we use the formula C (13,5). But we need to account for the fact that we do not know what particular suit we are going to use out of the 4, so we need to multiply C (13,5) * 4 to find all the possibilities for our five card flush. C (13,5) * 4 5148 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes. Self-critique: okay ------------------------------------------------ Self-critique rating: okay ********************************************* Question: `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since there are four suits there will be four different suit versions of 5,6,7,8, and 9. We know the ordered numbers we are looking for. Since we know for sure there are 4 different versions of these 5 particular numbers we can multiply 4 times 5. This will tell us how many straights there are. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: okay ------------------------------------------------ Self-critique rating: okay ********************************************* 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: To be honest with you I don’t know that much about cards - so when you say low or high I’m assuming you mean a low or high number. Which apparently must be wrong. I do not understand this question??? I went down to read the solution but I’m still really confused. I know where you are getting the 4^5 because we just worked that in one of the above questions but I don’t understand the low card bit. Where are you getting the 10 from??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights. STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one? INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card. Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To be honest with you I don’t know that much about cards - so when you say low or high I’m assuming you mean a low or high number. Which apparently must be wrong. I do not understand this question??? I went down to read the solution but I’m still really confused. I know where you are getting the 4^5 because we just worked that in one of the above questions but I don’t understand the low card bit. Where are you getting the 10 from??? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #*&!