course Mth 152 ¡wo—mʘë÷ÅÊеõñŸ®P|ב±´assignment #002
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21:41:03 `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 --> with replacement 26*26*26 Without replacement 26*25*24 confidence assessment: 2
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21:42:17 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 understand that this problem is asking for the number of words possible without replacement therefor my second answer of 26*25*24 is correct. self critique assessment: 2
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21:43:55 `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*26*25 confidence assessment:
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21:53:05 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'm not sure I follow this concept. Does the unordered collections the question refering too mean that abc, acb, bac, bca, cab, and cba considered the same and that is why there are only 1/6 the possiblites? Therefore, 26*25*24 divided by 6 would give you the possbillites. self critique assessment: 2
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21:55:49 `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 confidence assessment: 1
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22:01:08 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 left the second part out of my answer, but I don't understand where the 2 is coming from is this because of the number of digits to be chossen for each combincaiton is 2. I do understand that out of the 105 possiblites only 1 of those are 29 or more and therefore there are 104 possible combinations with a sum of less than 29 self critique assessment: 2
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22:26:19 `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 --> I'm not really sure on this one, but I'll take a stab at it. 48*47*46/2=51888 51888/26=1995.69 51888/7=7412.57 51888/15=3459.2 confidence assessment: 2
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22:32:31 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 didn't understand that they were in seperate bags. I thought that all of the bags contents had had been placed in one big box. I understand that there are 7 total rings so that makes 7 ways for the 1st bag and 6 for the 2nd and you divide 7*6 by 2 because each one could happen twice. self critique assessment: 2
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22:37:55 `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 --> confidence assessment:
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22:42:15 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 that you must subtact one from the total each time you take one from the bag ex if there are 26 tiles and you want to draw one then the next time you want to draw one there will only be 25 possiblities of drawing tile. Therefore, 15*26*25*7*14 possible ways to get them in that order. self critique assessment: 2
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22:43:02 `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*15*26*26*7 confidence assessment: 1
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22:57:05 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 --> I understand that once again if an object is drawn then there is one less possiblity that it will be drawn agian. I also understand that if there is no regard for order then there are have the possible outcomes. Thus, 15*14*26*25*7/(2*2) self critique assessment: 2
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22:58:36 `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 --> 15+26+7-7=5+26 confidence assessment: 1
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23:03:56 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 --> At first I meant to say 26-26+7+15=22 I understand that this is not complete and that you must say 22 ways of the 1st 21 for the 2nd ect. to get 22*21*20*19*18 self critique assessment: 2
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23:05:23 `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 --> 15*14*13*26*25 confidence assessment: 1
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23:06:20 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 --> Opps I forgot that the item was replaced therefore I should of answerd 15*15*15*26*26 self critique assessment: 2
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