course Mth152 õ«Zšèî}ŒY© ౿¬Ü ÈëVüa÷öDªíˆåüõïÎassignment #002
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23:17:38 `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 --> There are 26 letters in the alphabet so there would be 26 for the first pick, 25 for the second pick and 24 for the third pick. 26 x 25 x 24 =15,600 confidence assessment: 3
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23:17:48 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 --> o.k self critique assessment: 3
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23:29:31 `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 --> Going back to the previous question we had 26 x 25 x 24 = 15,600 Since the tiles are going to be unordered this time there are 6 ways the tile could be chosen. ex. jkl jk, jl, kl, kj, lj, lk So, then you would have to divide the ordered way 15,600 by the unordered 6 15,600/6=2,600 confidence assessment: 2
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23:29:43 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 --> o.k. self critique assessment: 3
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23:40:29 `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 x 14 = 210 There are two possible ways of getting each set of two balls out of the box ex: 14, 15 or 15, 14 so that cuts the 210 in half or 210/2 = 105 14 + 15 is the only combination that equals 29 so all the rest would be less than 29 confidence assessment: 2
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23:40:39 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 --> o.k. self critique assessment: 3
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23:57:54 `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 --> there are 7 total rings so the first bag could have 7 ways of having a ring. The second bag would only have 6 way. 7 x 6 = 42 There would be 2 different ways of coming up with each ring combination so that would be half of 42 or 21. 26 ways the first bag would have a tile x 25 ways the second bag would have a tile and again because of the two ways a combination could be made. 26 x 25 = 650 / 2 = 325 26 tiles x 7 rings divided by 2 possible combination ways = 91 Altogether there are 48 bags but 22 of the bags don't contain tiles. confidence assessment: 1
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00:02:05 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 was pretty certain of the first part of this question, but then I had a brain freeze on the very last part. I do see and understand now that I had to find out all of the combinations and subtract the the combinations that could not possibly have a tile. self critique assessment: 2
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00:07:43 `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 --> We have 15 balls so there are 15 chances of getting a ball on first pick. We still have all of the tiles so the second chance would be 26. For pick three we have already taken one of the tiles so the chance would be 25. We have all of the rings 7. We took one of the balls on the first pick so now there would only be 14 so we would have 15 x 26 x25 x 7 x 14 = confidence assessment: 2
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00:07:57 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 --> o.k. self critique assessment: 3
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