course Mth 152 I am having a hard time with this work. I have watched the CD and read the chapters but it seems that I cannot figure it out real good. I guess that is why I have been putting off doing the assignments is because I don't under it. Do you have any suggestions? tňMgƾ집assignment #002 002. Permutations, combinations, rearranging letters of words. Liberal Arts Mathematics II 02-09-2008 XRlݿפxβP assignment #002 002. Permutations, combinations, rearranging letters of words. Liberal Arts Mathematics II 02-09-2008
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10:48:44 `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 --> 26*25*24=15600 three letter words confidence assessment: 2
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10:49:16 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 --> That is how I worked it out getting 15600 self critique assessment: 2
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10:56:20 `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 --> since there are 3 ways the letters can be placed you would mutiply 3*2*1 = 6 and there are 26*25*24 choices you would multiply 26*25*24 and divide the answer by 6 = 2600 confidence assessment: 2
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10:56:51 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 --> That is how I worked it out also self critique assessment: 2
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N|Ʊհ assignment #002 002. Permutations, combinations, rearranging letters of words. Liberal Arts Mathematics II 02-09-2008
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15:16:20 `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 --> confidence assessment:
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15:16:24 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 --> self critique assessment:
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15:16:30 `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 --> confidence assessment:
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15:16:33 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 --> self critique assessment:
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15:19:47 `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 --> 15*14 =210 and there are two numbers 2, 9 so we would divide by 2 = 105. 105 possible confidence assessment: 2
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15:22:10 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 --> I forgot to take off the 14 + 15 that equals 29 so I got 105 instead of 104. self critique assessment: 2
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15:46:19 `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 --> 3 boxes - 1 box 15 balls; 1 box 26 tiles; 1 box 7 rings All mixed together 7 rings 26 tiles 26*7= 182 tile and ring 3*2*1 = 6 tiles confidence assessment: 2
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15:52:19 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 --> I had a very hard time with this problem. After I looked at your explanation I see how to work it out now. The balls would be divided by 2 because there are 2 ways that the numbers may come out. The tiles aren't divided by 2 becuase the alphabet would only come out as one letter and since there isn't as many balls as tiles you have to subtract to find the correct number. self critique assessment: 2
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16:05:42 `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 --> 15 balls, 26*25 tiles, 7 rings, 14 balls 15 + 650 + 7 + 14 =686 confidence assessment: 2
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16:07:36 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 added instead of mutiplying, but I got the right number of sets I just didn't mutiply. self critique assessment: 2
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16:15:47 `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 --> All together there would be 48 bags so if you subtract 26 bags of tiles there would be 22 bags left. 22*21*20*19*18 numbers and rings confidence assessment: 2
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16:16:40 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 --> I thought since there was five bags you would do the 22*21..... to get the answer self critique assessment: 2
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16:19:51 `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 --> If you are putting the numbers back in the box you would have 15 choices each time and 26 choices for tiles. since there are 3 numbers 15*15*15 and 2 letters 26*26 15*15*15*26*26 confidence assessment: 2
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16:20:34 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 --> I wasn't real sure about it but after reading the problem and realizing that the numbers and tiles were going back each time, I figured it out. self critique assessment: 2
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