course Mth 152 I am going to take this test on Monday. I hope that isn't a problem. Thank you. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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 understood this fairly well. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: Like the previous problem, you must do it similarly. For the 5’s and also the 9’s, you have the same way of obtaining them. That is they are chosen by C(4, 2) for both the 5’s and the 9’s. Since you have 4 cards, and there are 2 9’s left and 2 5’s left in the deck you have a total of 44 cards which are not 9’s or 5’s. Multiply them all out C(4,2) and C(4,2)* 44 is the amount of possible ways to get a hand with 2 5’s and 2 9’s. 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: This was easy to understand. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: Getting a full house for these numbers is C(4,2) for the 5’s and C(4,3) for the 9’s. You multiply this for C(4,2)*C(4,3) 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 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:I understood this as well. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: Just like the previous problem you use C(4, 2) for the two fives and C(4,3) for the face cards. However, these face cards could be a Jack, Queen, or King, so you have to multiply again by 3 to obtain the possible ways. 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 I understood this better after reading the explanation. The 3 identical face cards tripped me up a little bit. ------------------------------------------------ Self-critique rating: ********************************************* 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: To get the full house you have to have one set of numbers be a pair and then triple set of another number which is C(4,2)*C(4,3). There are 13 different types of cards in a deck (A, 2, 3, …, J, Q, K) so for the first choice you have 13. Getting three of another requires that the triple set is not a part of the pair (since it will also be a different hand) leaving 12 possible choices for the triple set. So you multiply 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 I was pretty confused when I was doing this problem and wasn’t sure if I had it correct but the solution cleared that up for me. ------------------------------------------------ Self-critique rating:2 ********************************************* 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: To get a flush you have to have to have 5 of the same suit cards and since there are 13 different types of cards, each in one of the 4 different suits, you have to have 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 this one was a bit easier to understand. ------------------------------------------------ Self-critique rating:3 ********************************************* 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 5’s, 6’s, 7’s, 8’s, and 9’s. In order to get a straight with these cards, you have to multiply the amount of 4’s in total, which is 4*4*4*4*4 for 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. Self-critique:I understood this as well. ------------------------------------------------ Self-critique rating:3 ********************************************* 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: Using an ace as either high or low to make a straight, there are 10 possible ways to start out the process (A, 2, 3, …10). So that gives 10. Like the previous problem, 5 was used as the low card giving 4*4*4*4*4 for the total of possibilities, so you must do the same and multiply this by 10 because there are 10 ways to start a straight if ace is used to fill the lowest or the highest spot. 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. I understood this but thanks for including the responses because I wasn’t sure whether or not to go about this like I did with the previous problem. "