course Mth 152 5:40 pm 9/13/09 If your solution to stated problem does not match the given solution, you should self-critique per instructions atvvvv
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Given Solution:n: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understood this well. ------------------------------------------------ Self-critique Rating:3 Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If we choose three letter tiles from the third box, then how many unordered collections of three letters are possible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since you are making three choices, and in an unordered collection, you would have to divide the 26*25*24 by 3*2*1 or coming out with 1/6 of the 15600 choices. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:n: If the 3-tile collections are unordered there are only 1/6 as many possibilities as for the ordered collection, since there are 3 * 2 * 1 = 6 orders in which any given collection might 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I could still use a bit of explanation as to why you do the 3*2*1 for an unordered collection. I understand about making the three choices but the 6 orders in choosing the given collection confuses me.
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Given Solution:n: The smallest possible total would be 1 + 2 = 3 and the greatest possible total would be 14 + 15 = 29. Thus the only way to get a total of 29 is to have chosen 14 and 15, in either order (i.e., either 14 first and 15 second, or 15 first and 14 second). 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess what I am unsure of when it comes to dividing these out by the possible choices is the way you arrive at 2 for this problem and 6 for the last one. I realize that you are making two choices each time, but for the last problem, you are only making three, then two then one and they are unordered, but for some reason its not clicking in my brain. ------------------------------------------------ Self-critique Rating:2 Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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 include two rings? How many of the possible combinations will include two tiles? How many of the possible combinations will include a tile and a ring? How many of the possible combinations will include at least one tile? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 7 rings, so 7 choices of rings for the first bag, and 6 for the second. Picking out two bags at random (in no particular order), leaves us to divide by the 2 choices made. 21 possible combinations for two rings. You would do the same for the 26 tiles. 26*25/2 or 325 possible combinations for the two tiles. For a tile and a ring you would use 26*7/2 for 91 possible combinations. For at least one tile, you have to imagine that there are 15 bags for the balls, 26 for the tiles, and 7 for the rings. This gives 48 bags in all. Only 26 of these contain a letter tile. 48-26 is 22. This leaves, for two choices, 22*21/2 for the number of bags without tiles= 231. Out of 48 bags, and you have two choices, 48*47/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:n: 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 at least one tile is therefore the difference 48 * 47 / 2 - 22 * 21 / 2 between the number of no-tile combinations and the total number of possible combinations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see how you arrived at this conclusion, but without looking at the answer, I was quickly confused. Looking over the correct way to go about this within the answer sort of cleared up the reason why I was confused. ------------------------------------------------ Self-critique Rating:2 Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Suppose we have mixed the contents of the three boxes as described in the preceding problem. If we pick five bags at random, then in how many ways can we get a ball, then two tiles in order, then a ring, then another ball, in that order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can get 15 balls on the first order, 26 tiles for the next choice, and 25 for the second getting two in a row, 7 for the ring, and 14 for the last ball if you go in that order. So using the order principal, you arrive at 15*26*25*7*14. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:n: There are five objects to choose. We apply the fundamental counting principle to each of the five choices: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This was much easier to grasp than the previous problem for some reason.
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Given Solution:n: This time the order in which the choices are made doesnt matter. 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 choseb 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 make an unordered choice of 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand how to make the proper number of choices when it comes to order/and unordered, but flipping it and going by the first part of your solution confuses me slightly. ------------------------------------------------ Self-critique Rating:2 Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 26 bags that contain a tile and 22 that do not. If you are choosing 5 bags and they do not contain a tile, you come up with 22*21*20*19*18/120?=26334 You are making 5 choices, so you must divide by 5*4*3*2*1 if my understanding is correct. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:n: Of the 48 bags, 26 do and 22 do not contain a tile. If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ordered ways in which the five bags could all contain something besides a tile. Any given collection of five bags could have been chosen in any of 5 * 4 * 3 * 2 * 1 orders. There are therefore 22 * 21 * 20 * 19 * 18 / (5 * 4 * 3 * 2 * 1) possible unordered collections of five bags. GOOD STUDENT COMMENT: Ok, so in the previous questions, when deciding what number to divide by, you always multiply the number of orders that a bag could have been chosen. So when dividing by 2, it was because you were only choosing two bags (2*1=2). But in this case, there are 5 bags to choose from (5*4*3*2*1). Then you divide the number of ordered ways by this number. INSTRUCTOR RESPONSE: Very good. You stated that well. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, after seeing the student comment and the solution I think I have a better grasp on how to divide when order doesnt matter. ------------------------------------------------ Self-critique Rating:2 Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the order matters here and you are replacing the ball, you have 15 choices for each time you choose a ball (15*15*15) and the same happens with the two tiles, for 26*26. This gives a total of 15*15*15*26*26=2281500 possible words that can be 3 numbers and 2 letters that are formed from the result. You are not losing choices each time you choose a tile or a ball because you are replacing it, and can therefore choose it again. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:n: 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'. "