#$&* course Mth 152 6/21 7:30 002. Permutations, combinations, rearranging letters of words.
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Given Solution: There are 26 choices for the first tile chosen, 25 for the second and 24 for the third. The number of possible three-letter words with 3 distinct letters of the alphabet is therefore 26 * 25 * 24. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. If we choose three letter tiles from the second box, then how many unordered collections of three letters are possible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If your choosing three tiles again yet the choices are not ordered, you would have 3*2*1=6 possibilities . 26*25*24/6=2600 possibilities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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 it’s 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: 14+15= 29. We are looking for all of the totals with less than 29 as the total added up. In order to get the answer it is easier to first get the total of times that the balls will be about or equal to 29. 15*14/2= 105 combinations equal or greater than 29. Therefore there are a remaining amount of 104 balls that could be selected that are less than 29. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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: Since there are only 7 rings, there are only 7 different ways for the first bag, 6 for the second, 5 the third and so on til the last ring. Since we are looking for two rings, there are two different orders the rings could be bagged. 7*6/2=21 ways of getting two rings per bag. There are 26 tiles to choose from and of course 7 rings still with two in each. 26*7/2= 91 combinations of the two bags, one with a ring and one with a tile. There would be a total of 48 bags simply because of the amount of objects you have need to be added to each bag. 15 balls+ 26 tiles+ 7 rings= 48 bags. The total for the two bag combinations would be 48*47/2=1128. There would be 22 bags not containing rings because 15+7=22. Therefore, you would have 22*21/2=231 bags without tiles in them. You must find the difference of the total of tile containing bags and non-tile containing bags. (48*47/2)-(22*21/2)=1128-231=897 possible bags containing each object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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 have a total of 5 objects to choose. There are 15 possible ball choices, 26 possible tile choices for the second selection and one less for the third selection making that 25. The fourth choice has 7 different colored rings to choose. And on the fifth choice you would have 14 balls because of the first choice. Therefore it would be 15*26*25*7*14=955500 for the total possibility of items per bag. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are five objects to choose. We apply the fundamental counting principle to each of the five choices: •There are 15 bags containing balls, so there are 15 ways to get a ball on the first selection. •If a ball is chosen on the first selection, there are still 26 bags containing tiles when the second selection is made. So there are 26 ways to get a tile on the second selection. •At this point there are 25 tiles so there are 25 ways to get a tile on the third selection. •There are still 7 rings from which to select, so that there are 7 ways the fourth choice can be a ring. •Since 1 ball has been chosen already, there are 14 ways that the fifth choice can be a ball. To get the specified choices in the indicated order, then, there are 15 * 26 * 25 * 7 * 14 ways. STUDENT COMMENT This was much easier to grasp than the previous problem for some reason. INSTRUCTOR RESPONSE That's because you're thinking about the problems and asking good questions. This is the key to success. Keep up the good work. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. Suppose we have mixed the contents of the three boxes as in the preceding problems. If we pick five bags at random, then in how many ways can we get two balls, two tiles and a ring in any order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This problem would be just like the other except that there are two balls, two tiles, and a ring. (15*14/2)*(26*25/2)*7= 105*325*7=238875 possible outcomes confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This time the order in which the choices are made doesn’t 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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: For the total of 48 bags, there will be 26 that do contain a tile and 22 that do not contain a tile. Choosing five bags at random with order it would be 22*21*20*19*18. Since you are choosing from five different bags the combination would be 5*4*3*2*1. Therefore, there are (22*21*20*19*18)/(5*4*3*2*1)= 3160080/ 120= =26334 bags not containing a tile. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Of the 48 bags, 26 do and 22 do not contain a tile. •If we pick five bags at random, then there are 22 * 21 * 20 * 19 * 18 ordered ways in which the five bags could all contain something besides a tile. •Any given collection of five bags could have been chosen in any of 5 * 4 * 3 * 2 * 1 orders. •There are therefore 22 * 21 * 20 * 19 * 18 / (5 * 4 * 3 * 2 * 1) possible unordered collections of five bags. GOOD STUDENT COMMENT: Ok, so in the previous questions, when deciding what number to divide by, you always multiply the number of orders that a bag could have been chosen. So when dividing by 2, it was because you were only choosing two bags (2*1=2). But in this case, there are 5 bags to choose from (5*4*3*2*1). Then you divide the number of ordered ways by this number. INSTRUCTOR RESPONSE: Very good. You stated that well. Self-critique: This problem was confusing for me simply because of the ordered ways being divided by the number of bags. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q008. Suppose the balls, tiles and rings are back in their original boxes. If we choose three balls, each time replacing the ball and thoroughly mixing the contents of the box, then two tiles, again replacing and mixing after each choice, how many 5-character 'words' consisting of 3 numbers followed by 2 letters could be formed from the results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 15 balls to choose from, so if there is replacement there will be 15 outcomes for every ball. There are 26 possible outcomes for every choice of a tile with replacement as well. We are choosing three balls and two tiles so it would be 15 * 15 * 15 * 26 * 26 possible words. A total of 2281500 selections. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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'." Self-critique (if necessary): ------------------------------------------------ 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, how many 5-character 'words' consisting of 3 numbers followed by 2 letters could be formed from the results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 15 balls to choose from, so if there is replacement there will be 15 outcomes for every ball. There are 26 possible outcomes for every choice of a tile with replacement as well. We are choosing three balls and two tiles so it would be 15 * 15 * 15 * 26 * 26 possible words. A total of 2281500 selections. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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'." Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!