#$&* course Mth 152 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's. Self-critique: I was right I just didn’t explain it exactly like you did ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4c2*4c2*44 Four because that is the numbers in the set and two because that’s how many numbers you want and when you subtract those cards from the deck you end up with 44 cards so you multiply all of those together leaving you with the answer 1,584 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ 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: 4c2*4c3 because there are a total of four fives in a deck and you only need two and then there are four nines in a deck and you need three of them. You multiply these together and you are your answer 24 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3). Self-critique: I had to look at your answer because I’m confused as to when you multiply by the left over deck cards. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You use the same numbers from the last question. It’s the same difference with three identical face cards. 4c2*4c3*3, you multiply by 3 because of the three different face cards, giving you the answer 72 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ Self-critique rating:3 ********************************************* 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: 13*12*4c2*4c3 13 because there are 13 cards for a match 12 because then there is only 12 cards left for a match confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses. Self-critique: ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 13c5 because there are 13 cards in a suit and you only need five then multiply that by four because there are four suit all together confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ 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: You can do 4c1 for all other these and then multiply them together or you can just do 4^5 and get the answer. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9. STUDENT QUESTION not sure I understand why is it not C(20,5) I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not… INSTRUCTOR RESPONSE There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations. However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts. That's not a straight, nor are most of the C(20, 5) combinations of these cards. C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).< Self-critique: ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4^5 is how you figure out one straight however there are ten possible straight so you multiply 4^5*10 to get your answer of 10240 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 12c5 12 because there are 12 face cards in a deck and 5 because you only need five face cards in your hand therefore the answer is 792 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. Using a standard deck, in how many ways is it possible to get a 5-card hand which includes exactly two face cards? (Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12c2*40c3 12 because that’s how many face cards are in a deck and 2 because you only need two. 40 because that’s the remaining cards left of the deck and 3 because you only need three more cards to complete you hand. Giving you the answer of 652080 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: