#$&* course MTH 152 3:35 PM , 7/11/13 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
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: Seems a bit high of a number to get just two 5’s from a set of 52 cards…Is my solution correct?
.............................................
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: In my solution, am I correct in my hypothesis? Given that the 6 ways to get a 5, 6 ways to get a 9 = 52-12 = 40 cards remaining that have neither 5 or 9? I think I just answered my own question. I was reading the number of ways to get the card 5 or 9, not the amount of cards, being 4. That way, there are 4 + 4 = 8, 52 - 8 = 44 cards left that are neither 5 or 9, so therefore that amends my previous solution to 1584 possible ways. ------------------------------------------------ Self-critique rating: Ok. See my comment above. ********************************************* 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: - I have to check to see what a “full house” will consist of, to aid my understanding. - Apparently a full house is any hand of 5 cards in which there are 3 matching cards of one rank and another pair of cards with another rank. - So I would hypothesize that an example of a full house would be something like - [ 8(clubs), 8(spades), 8(diamonds), 5(hearts), 5(spades) ] or something to that effect. - So with this understanding of a full house being any 5-card combination where 3 cards will be of similar value but differing suit, and 2 cards of same value of another but similar suit. - The hand in question is two 5’s and three 9’s. - That means that being dealt two 5’s will be C(4, 2) = 6 ways to get a hand with two 5’s - Being dealt three 9’s can be read as C(4, 3) = 4! / [ (4 - 3)! * 3! ] = 4 * 3 * 2/ 3 * 2 * 1 = 2 * 1 * 2 = 4. - So now I consider that C(4,2) * C(4, 3) * 44 = 1056 ways to get a full house consisting of two 5’s and three 9’s. - *** - I need to change my answer because five cards are already chosen, not 4. - With this change there are approximately 6 * 4 = 24 ways to get a full house with two 3’s and three 9’s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. Getting the hang of this new kind of problem; perhaps the following sections will help me to count cards so I can win a fortune in Nevada. - Just kidding ------------------------------------------------ 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: - The first thing I fence with is the notion of getting two 5’s, which was addressed in the previous problems in such a way that it becomes second-nature. - C (4, 2) = 6 ways to be dealt a 5 from each of the four suits. - There are three different kinds of face cards, being Jack, Queen, and King. - C(4, 3) = 4 ways to get a face card (being a collection of 4 suits with 3 possible cards to choose from) - Then finally multiply the products of C(4,2) * C(4, 3) * 3 to result in = 72 ways to be dealt a full house with two 5’s and three identical face cards. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3). Self-critique: Why is the 3 multiplied here? I’m trying to get this in my head so I will know for future reference and present clarity. My hypothesis is that multiplying by 3 is: Suppose we are dealt [ 5(clubs), 5(spades), J(clubs), J(spades), J(diamonds)] C(4, 2) * C( 4, 3) * 3 = “the number of ways one can be dealt two 5’s and three J’s multiplied by the number of suits left for that card” to account for one identical card in question, say the one described in c(4,3) refers to the J(clubs) one would need to account for J(spades), J(hearts), and J(diamonds). ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - By “two of one denomination” I’m taking this to be synonymous with identical. - And by “three of another” I’m taking it to read 3 more cards of each different suit. - The constraint by a “full house” refers that the cards in question need to be identical in value but can be different suits. - So thus I begin by: - C(4,2) = 6 - C(4, 3) = 4 - C(4,2) * 2 * C(4,3) * 3 = 12 * 12 = 144 different ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 on the right track with my train of thought. I can see how you utilize the FCP (in a sense) of deducing that there are 13 possibilities for two pairs (two pairs meaning 1 and 1, then 1 and 1 =4 therefore 13 * 4 = 52, the entire deck) leaving 12 possibilities for the other two pairs (meaning 12 * 4 = 48. 52 - 4 = 48 cards remaining.) - Thus - - [ 13 * C(4,2) ] * [ 12 * C(4,3) ] = 3744 ways to get a pair of one suit and a three-of-a-kind for another. ***** - After taking a second-glance at this problem, wouldn’t there be 11 options left for the three-of-a-kind, since one pair would result in 13 - 2 = 11?
.............................................
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: Good to go for this one. Beginning to make connections. ------------------------------------------------ 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 13 cards in a suit. Only one suit will be considered because of the constraints listed as a “straight’ hand that is successive in order through descending/ascending order of the same suit. - I begin hypothesizing about getting specific cards, not just any random card, and how this would change the mathematical formula. - I could say that: C(13, 5) * 13 * 12 * 11 * 10 * 9 = the number of ways one could be dealt a straight consisting of specific denominations (but, this is a rather large number, being 90,347,400) - I should note that there are 4 suits to be considered, and since a specific suit was not mentioned, I should consider all 4. Thus being 4 * 4 * 4 * 4 * 4 (4^5) or 1024. But I was close. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I have to laugh at my first estimation, but I often second-guess myself for each answer I give anyway, especially when it’s too large a number. Regardless, I can see how this would be 4^5. *******My hypothesis is that for each of the five cards, there are four different suits (spades, clubs, diamonds, hearts) and apparently the denomination doesn’t really matter; I’m using the FCP in the wrong places. When can I use the FCP and be confident that its use is pertinent to the problem at hand?******
.............................................
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: - 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: