course Mth 152 ßôgí·¤¤¬ÜÆy±Êº‘Έ¿y®Ò…“assignment #002
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09:23:23 `q001. Note that there are 8 questions in this assignment. If we choose three letter tiles from a complete set, which consists of one tile for each letter, then how many three-letter 'words' are possible ?
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RESPONSE --> So we take 26*25*24=15600 Next we know that they can be in any order so 3*2*1=6 This gives us 15600/6=2600 confidence assessment: 3
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09:24:29 There are 26 choices for the first tile chosen, 25 for the second and 24 for the third. The number of possible three-letter words with 3 distinct letters of the alphabet is therefore 26 * 25 * 24.
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RESPONSE --> I answered this correctly but I also divided 26*25*24 by 6 because they could be in any order, was this wrong? self critique assessment: 2
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09:26:03 `q002. If we choose three letter tiles from a complete set, then how many unordered collections of three letters are possible?
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RESPONSE --> 26*25*24=15600 We have 6 possible collections from the 3 tiles selected so 15600/6=2600 confidence assessment: 3
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09:26:20 If the 3-tile collections are unordered there are only 1/6 as many possibilities, since there are 3 * 2 * 1 = 6 orders in which any collection could have been chosen. Since there are 26 * 25 * 24 ways to choose the 3 tiles in order, there are thus 26 * 25 * 24 / 6 possibilities for unordered choices.
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RESPONSE --> I answered this correctly self critique assessment: 3
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09:33:48 `q003. If we choose two balls from fifteen balls, numbered 1 - 15, from the first box of the preceding problem set, and do so without replacing the first ball chosen, we can get totals like 3 + 7 = 10, or 2 + 14 = 16, etc.. How many of the possible unordered outcomes give us a total of less than 29?
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RESPONSE --> I am not sure how to do this problem so I am moving on to the answer to better understand. confidence assessment: 0
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09:36:31 The smallest possible total would be 1 + 2 = 3 and the greatest possible total would be 14 + 15 = 29. We quickly see that the only way to get a total of 29 is to have chosen 14 and 15, in either order. Thus out of the 15 * 14 / 2 = 105 possible unordered combinations of two balls, only one gives us a total of at least 29. The remaining 104 possible combinations give us a total of less than 29. This problem illustrates how it is sometimes easier to analyze what doesn't happen than to analyze what does. In this case we were looking for totals less than 29, but it was easier to find the number of totals that were not less than 29. Having found that number we easily found the number we were seeking.
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RESPONSE --> OK I understand now. That was actually very simple. I was to busy concentrating on the numbers less than 29. If I would have tried to find out what number equal 29 the answer would have been clear. self critique assessment: 2
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09:45:50 `q004. If we place each object in all the three boxes (one containing 15 numbered balls, another 26 letter tiles, the third seven colored rings) in a small bag and add packing so that each bag looks and feels the same as every other, and if we then thoroughly mix the contents of the three boxes into a single large box before we pick out two bags at random, how many of the possible combinations will have two rings? How many of the possible combinations will have two tiles? How many of the possible combinations will have a tile and a ring? How many of the possible combinations will include a tile?
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RESPONSE --> Ok so if we are combining all the items into one box that gives us 26+15+7=48. So the large box will contain 48 items. If we pull two from the box we have 48*47=2256/2=1125. This is where I am unsure where to go next to answer the next questions. I am going to move on to the answer from here to see what I should do next. how many of the possible combinations will have two rings? How many of the possible combinations will have two tiles? How many of the possible combinations will have a tile and a ring? How many of the possible combinations will include a tile? confidence assessment: 2
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09:48:38 There are a total of 7 rings. There are thus 7 ways the first bag could have contained a ring, leaving 6 ways in which the second bag could have contained a ring. It follows that 7 * 6 / 2 = 21 of the possible combinations will contain 2 rings (note that we divide by 2 because each combination could occur in two different orders). Reasoning similarly we see that there are 26 ways the first bag could have contained a tile and 25 ways the second bag could have contained a tile, so that there are 26 * 25 / 2 = 325 possible combinations which contain 2 tiles. Since there are 26 tiles and 7 rings, there are 26 * 7 / 2 = 91 possible two-bag combinations containing a tile and a ring. There are a total of 15 + 26 + 7 = 48 bags, so the total number of possible two-bag combinations is 48 * 47 / 2. Since 15 + 7 = 22 of the bags do not contain tiles, there are 22 * 21 / 2 two-bag combinations with no tiles. The number of possible combinations which do include tiles is therefore the difference 48 * 47 / 2 - 22 * 21 / 2 between the number of no-tile combinations and the total number of possible combinations.
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RESPONSE --> Ok I see how this works now. self critique assessment: 3
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10:00:26 `q005. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get a ball, then two tiles, then a ring, then another ball, in that order?
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RESPONSE --> Since we only have 7 rings and we know that we must have one ring in our set of a ball, then two tiles, then a ring, then another ball, in that order then we know we could only have this combination 7 times because once we run out of rings the combination could no longer continue confidence assessment: 2
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10:03:09 There are 15 bags containing balls, so there are 15 ways to get a ball on the first selection. If a ball is chosen on the first selection, there are still 26 bags containing tiles when the second selection is made. So there are 26 ways to get a tile on the second selection. At this point there are 25 tiles so there are 25 ways to get a tile on the third selection. There are still 7 rings from which to select, so that there are 7 ways the fourth choice can be a ring. Since 1 ball has been chosen already, there are 14 ways that the fifth choice can be a ball. To get the specified choices in the indicated order, then, there are 15 * 26 * 25 * 7 * 14 ways.
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RESPONSE --> I understand this. I would have done this but I misunderstood the question. I was answering how many times this could happen and since there were only 7 rings I answered that this combination could only happen 7 times. My mistake self critique assessment: 3
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10:07:45 `q006. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get two balls, two tiles and a ring in any order?
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RESPONSE --> 15*14*26*25*7=955500 Since order is not an issue we need to find out how many possible combinations we have. We do this by 5*4*3*2*1=120. This gives us 955500/120=7962.5 confidence assessment: 2
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10:10:50 There are 15 * 14 possible outcomes when 2 balls are chosen in order, and 15 * 14 / 2 possible outcomes when the order doesn't matter. There are similarly 26 * 25 / 2 possible outcomes when 2 tiles are choose without regard for order. There are 7 possible choices for the one ring. Thus we have [ 15 * 14 / 2 ] * [ 26 * 25 / 2 ] * 7 ways in which to choose 2 balls, 2 tiles and a ring. Another way to get the same result is to start with the 15 * 26 * 25 * 7 * 14 ways to choose the 2 balls, 2 tiles and one ring in a specified order, as shown in the last problem. Whichever 2 tiles are chosen, they could have been chosen in the opposite order, so if the order of tiles doesn't matter there are only half as many possible outcomes--i.e., 15 * 26 * 25 * 7 * 14 / 2 possibilities if the order of the tiles doesn't matter that the order of the balls does. If the order of the balls doesn't matter either, then we have half this many, or 15 * 26 * 25 * 7 * 14 / ( 2 * 2) ways. It should be easy to see why this expression is identical to the expression [ 15 * 14 / 2 ] * [ 26 * 25 / 2 ] * 7 obtained by the first analysis of this problem.
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RESPONSE --> Ok I am really having a problem with this section. I understand alot of the process but I am having trouble on when to use what methods. After looking at the answers it all is really simple and I know the process. I am just confused on when to use which process. self critique assessment: 1
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10:12:36 `q007. Suppose we have mixed the contents of the three boxes as described above. If we pick five bags at random, then in how many ways can we get a collection of objects that does not contain a tile?
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RESPONSE --> I am moving on to the answer. I believe I know how to answer the question but again I am unsure if I am going to use the right method. I do not want to confuse myself so I will move on to the answer. confidence assessment: 0
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10:14:57 Of the 48 bags, 22 do not contain a tile. If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ways in which the five bags could all contain something besides a tile.
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RESPONSE --> Ok I was actually thinking this was the method. I was going to minus the number of tiles and then do 22*21*20*19*18. I completely understand this question but I am still unsure when to use each method. self critique assessment: 3
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10:15:02 Of the 48 bags, 22 do not contain a tile. If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ways in which the five bags could all contain something besides a tile.
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RESPONSE --> self critique assessment:
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10:18:34 `q008. Suppose the balls, tiles and rings are back in their original boxes. If we choose three balls, each time replacing the ball and thoroughly mixing the contents of the box, then two tiles, again replacing and mixing after each choice, then how many 5-character 'words' consisting of 3 numbers followed by 2 letters could be formed from the results?
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RESPONSE --> 26*26/2 * 15*15/2=38025 confidence assessment: 1
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10:20:21 Since the order of the characters makes a difference when forming 'words', the order of the choices does matter in this case. We have 15 balls from which to choose, so that if we choose with replacement there are 15 possible outcomes for every choice of a ball. Similarly there are 26 possible outcomes for every choice of a tile. Since we first choose 3 balls then 2 tiles, there are 15 * 15 * 15 * 26 * 26 possible 5-character 'words'.
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RESPONSE --> Ok I forgot to do 15 3 times at we choose the balls 3 times. I did it two times but I understand the method. My only question is would you not divide the balls by 3 and the tiles by 2. Why would you not do that for this question? self critique assessment: 2
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