#$&* course Mth 152 1/24 1:10 PM Question: `q001. Note that there are 8 questions in this assignment.
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Given Solution: There are 26 choices for the first tile chosen, 25 for the second and 24 for the third. The number of possible three-letter words with 3 distinct letters of the alphabet is therefore 26 * 25 * 24. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If we choose three letter tiles from the third box, then how many unordered collections of three letters are possible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the 3-tile collections are unordered there are 1/6 as many possibilities as for ordered collections there are 3*2*1 = 6 so we divide that by 26*25*24 to get possible unordered collections. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: 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. STUDENT QUESTION 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. INSTRUCTOR COMMENT For example you could choose the c, m and q tiles in any of the six orders cmq, cqm, mcq, mqc, qcm, qmc. If you are making ordered choices, all six would be different. But if choices are unordered, they all contain the same letters so these six ordered choices would all be the same. Clearly the same is true for any collection of 3 letters. So there are 6 times more ordered choices than unordered choices. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. If we choose two balls from from the first box of the original problem, 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? STUDENT QUESTION 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. INSTRUCTOR RESPONSE If you choose balls 3 and 7, then they could be chosen in the order 3,7 or in the order 7,3. The unordered collection is the same, no matter which ball is chosen first. So a single unordered collection of two objects corresponds to 2 different ordered choices. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The smallest possible total would be 1+2=3 and the greatest would be 14+15=29 and the only way to get 29 is to have chosen 14+15 so therefore 14*15/2 = 105 possible unordered combinations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: 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 chosen14 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): ------------------------------------------------ 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 boxbefore 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 seven ways the 1st bag could have a ring; 6 ways the 2nd bag could have a ring; so therefore we use 7*6/2 = 21 possible combinations for 2 rings 26 ways the 1st bag could have a tile 25 ways the 2nd could have a tile so 26*25/2 = 325 possible for 2 tiles .. now 26*7/2 = 91 possible for 2 bag combinations of a tile and ring and a total of 48 bags so 48*47/2 and 48*47/2 -22*21/2 equals the difference for bags that have one tile. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: 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): ------------------------------------------------ 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: We use the Fundamental Counting Principle for each of the 5 choices, getting 15 for the 1st bag, 26 for the 2nd, 25 for the 3rd, 7 for the 4th, and 14 for the 5th so therefore we have 15*26*25*7*14 ways. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are five objects to choose. We apply the fundamental counting principle to each of the five choices: There are 15 bags containing balls, so there are 15 ways to get a ball on the first selection. If a ball is chosen on the first selection, there are still 26 bags containing tiles when the second selection is made. So there are 26 ways to get a tile on the second selection. At this point there are 25 tiles so there are 25 ways to get a tile on the third selection. There are still 7 rings from which to select, so that there are 7 ways the fourth choice can be a ring. Since 1 ball has been chosen already, there are 14 ways that the fifth choice can be a ball. To get the specified choices in the indicated order, then, there are 15 * 26 * 25 * 7 * 14 ways. STUDENT COMMENT This was much easier to grasp than the previous problem for some reason. INSTRUCTOR RESPONSE That's because you're thinking about the problems and asking good questions. This is the key to success. Keep up the good work. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. Suppose we have mixed the contents of the three boxes as in the preceding problems. If we pick five bags at random, then in how many ways can we get two balls, two tiles and a ring in any order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 15*14/2 multiplied by 26*25/2 and multiply 7 ways to get the unordered choices of 2 balls, two tiles, and a ring. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: 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): ------------------------------------------------ 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: Of the 48 bags, 26 do contain and 22 do not contain a tile. So there are 22*21*20*19*18 ordered ways. Any given collection of 5 bags (5*4*3*2*1 for ordered ) So therefore we use 22*21*20*19*18/5*4*3*2*1 to equal the possible unordered collections of 5 bags. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Of the 48 bags, 26 do and 22 do not contain a tile. If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ordered ways in which the five bags could all contain something besides a tile. Any given collection of five bags could have been chosen in any of 5 * 4 * 3 * 2 * 1 orders. There are therefore 22 * 21 * 20 * 19 * 18 / (5 * 4 * 3 * 2 * 1) possible unordered collections of five bags. GOOD STUDENT COMMENT: Ok, so in the previous questions, when deciding what number to divide by, you always multiply the number of orders that a bag could have been chosen. So when dividing by 2, it was because you were only choosing two bags (2*1=2). But in this case, there are 5 bags to choose from (5*4*3*2*1). Then you divide the number of ordered ways by this number. INSTRUCTOR RESPONSE: Very good. You stated that well. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Suppose the balls, tiles and rings are back in their original boxes. If we choose three balls, each time replacing the ball and thoroughly mixing the contents of the box, then two tiles, again replacing and mixing after each choice, then how many 5-character 'words' consisting of 3 numbers followed by 2 letters could be formed from the results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We choose 3 balls then 2 tiles,so there are 15 * 15 * 15 * 26 * 26 possible 5-character 'words'. confidence rating #$&*: ********************************************* 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: We choose 3 balls then 2 tiles,so there are 15 * 15 * 15 * 26 * 26 possible 5-character 'words'. confidence rating #$&*: #(*!