course Mth 152 ܱơEy甁R¡xassignment #005
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10:35:58 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 --> 2^6 -6 = 58 Make two dice and test it.
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10:37:27 ** 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 computed 2^6 instead of 6^2.
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10:39:30 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> ?
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10:40:19 ** 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 do not have this problem in my book.
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10:42:00 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 --> 36. 2^12 - 2^5 = 4064
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10:42:56 ** 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 --> ok
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10:50:47 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 --> ?
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10:51:29 ** 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 --> ok
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10:56:25 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE --> Ends in 25 or 50. 4 numbers can end up divisible by 25. 5 numbers are divisible by 50. 4+5 = 9
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10:58:43 ** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities. A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25. A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650. Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement. The fundamental counting principle is the key here. STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations. You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
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RESPONSE --> ?
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11:00:34 Query 11.5.48 # 3-digit counting #'s without digits 2,5,7,8?
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RESPONSE --> 5*6*6 = 180
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11:00:37 ** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
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RESPONSE --> ok
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11:00:59 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The problems still do not match up with my book.
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