course Mth 152 ???r??{+??????assignment #005005. `query 5 Liberal Arts Mathematics II 02-23-2008
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21:21:28 Query 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
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RESPONSE --> 6 * 6 = 36 If I was marrooned, I would work it out by pairing the numbers together.
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21:22:24 ** On two fair dice you have 6 possible outcomes on the first and 6 on the second. By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes. We can list these outcomes in the form of ordered pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Of these 36 outcomes there are six that have the same number on both dice. It follows that the remaining 3 - 6 = 30 have different numbers. So there are 30 ways to get different numbers on the two dice. Note that your chance of getting different numbers is therefore 30 / 36 = 5/6 = .8333... .**
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RESPONSE --> I did not subtract the 6 from the 36.
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21:30:04 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> I'm sorry but I do not understand this question. I'm not sure what a suit is so I am going to go to your answer to see how it works. Sorry
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21:34:20 ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13). There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit). Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands. Of the number of hands having more than one suit is C(52, 13) - 4. **
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RESPONSE --> I know now from your explanation how to work the problem. You find out how many suits and then you subtract
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22:03:36 11.5.20 # subsets of 12-elt set with from 3 to 9 elts? How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
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RESPONSE --> C(12,0) = 1 C(12,1) = 12 C(12,2) = 12! / (2! (12-2)!) = 12! / (2! * 10!) = 12*11 / 2*1 = 132 / 2 = 66 C(12,10)= 12! / (10! (12-10)!) = 12! / (10! * 2!) = 12*11 / 2*1 = 132/2 = 66 C(12,11)= 12! / (11!(12-11)!) = 12! / (11! * 1!) = 12 / 1 = 12 C(12,12) = 0 2^12 = 2*2*2*2*2*2*2*2*2*2*2*2 = 4096 - (1+12+66+66+12=0) = 4096 - 157 = 3939
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22:04:27 ** You need the number of subsets with 3 elements, with 4 elements, etc.. You will then add these numbers to get the total number of 3-, 4-, 5-, ., 9-element subsets. Start with a 3-e.ement subset. In a 12-element set, how many subsets have exactly three elements? You answer this by asking how many possibilities there are for the first element, then how many for the second, then how many for the third. You can choose the first element from the entire set of 12, so you have 12 choices. You have 11 elements from which to choose the second, so there are 11 choices. You then have 10 elements left from which to choose the third. So there are 12 * 11 * 10 ways to choose the elements. However, the order of a set doesn't matter. 3 elements could be ordered in 3! different ways, so there are 12 * 11 * 10 / 3! ways to choose different 3-element sets. This is equal to C(12,3). So there are C(12, 3) 3-elements subsets of a set of 12 elements. Reasoning similarly we find that there are C(12,4) ways to choose a 4-element subset. C(12,5) ways to choose a 5-element subset. C(12,6) ways to choose a 6-element subset. C(12,7) ways to choose a 7-element subset. C(12,8) ways to choose a 8-element subset. C(12,9) ways to choose a 9-element subset. We see that there are C(12,3) + C(12,4) + C(12,5) + C(12,6) + C(12,7) + C(12,8) + C(12,9) possible subsets with 3, 4, 5, 6, 7, 8 or 9 elements. Alternatively you can figure out how many sets have fewer than 3 or more than 9 elements. There are C(12, 0) + C(12, 1) + C(12, 2) = 1 + 12 + 66 = 79 sets with fewer than 3 elements, and C(12, 10) + C(12, 11) + C(12, 12) = 66 + 12 + 1 = 79 sets with more than 3. Since there are 2^12 = 4096 possible subsets of a 12-element set there are 4096 - 79 - 79 = 3938 sets with between 3 and 9 elements. **
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RESPONSE --> That is how I worked it, I got all the subsets not included and subtracted them from the total subsets.
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22:12:43 11.5.30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
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RESPONSE --> C(52,5) 52! / (5!(52-5)!) = 52! / (5! * 47!) = 52*51*50*49*48 / 5*4*3*2*1 = 311875200 / 120 = 2598960 2598960 - 2588760 = 10200
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22:14:46 ** There are 9 choices for the denomination of the lowest card in a straight, which gives 36 cards that could be the low card. However but if aces can be high or low there are 40. There are then four choices for the next-higher card, four for the next after that, etc., giving 40*4*4*4*4 possibilities. **STUDENT COMMENT: I don 't understand this one . Idon't see where you get the 9 from. INSTRUCTOR RESPONSE: Cards run from 2 through 10, then the four face cards, then the ace. You need five consecutive cards to make a straight. The highest possible straight is therefore 10, Jack, Queen, King and Ace. The lowest is 2, 3, 4, 5, 6. The lowest card of the straight can be any number from 2 through 10. That is 9 possibilities. In some games the ace can be counted as the low card, below the 2, as well as the high card. In that case there would be one more possibility for a straight, which could then consists of denominations 1, 2, 3, 4, 5. *&*&
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RESPONSE --> I see where you got the answer you have to work throught it and figure out what isn't included and then work the problem.
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22:17:11 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE -->
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