#$&* course MTH 152 Question: `q001. Note that there are 9 questions in this assignment.
.............................................
Given Solution: There are 26 choices for the first tile chosen, 25 for the second and 24 for the third. The number of possible three-letter words with 3 distinct letters of the alphabet is therefore 26 * 25 * 24. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My first time through was spot on, but I didn’t feel confident when I saw how large the solution was and then second-guessed myself. ------------------------------------------------ Self-critique rating: Ok, but see my comment above and provide any insight where appropriate. ********************************************* Question: `q002. If we choose three letter tiles from the second box, then how many unordered collections of three letters are possible? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 2600. Taking from the solution above, 15600, dividing by 3 * 2 * 1 = 6 - Should one make a note to reference that whenever an “unordered collection” is given in a problem that one should automatically think “divide by the product of the number of items contained in the unordered collection in a descending fashion?” YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Shouldn’t the unordered values be more numerous than the ordered values? If not, why? Elaborate further or perhaps paraphrase what you mean by “ordered” and “unordered”
.............................................
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): Oftentimes I read “too far” into the wording of a problem and cannot comprehend exactly what is asked. For example, taken from your opening statement about problem 3, “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.. “ The portion of the statement outlined by asterisks is what led me to believe that one could not change the first ball, so naturally the rest of the question quickly fell apart. However, after reading below to the Given Solution to find out what’s going on, I caught on fairly briskly and followed every step. As such, I paraphrase these kinds of situations to further cement my understanding. You will know by my extensive paraphrases of a question I don’t understand, so I list out every thought process for you to critique later. ------------------------------------------------ Self-critique Rating: Ok, see my comments in bold font above. ********************************************* 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: How many possible combinations will include 2 rings? - First I think how many rings are contained within the third box, which is 7. - Second I think how many times one will select a bag at random, which is 2 times. - Thus, 7 * 6 = 42, then 42/2 = 21. - Therefore, 21 possible combinations will include two rings. How many possible combinations will include 2 tiles? - First I remind myself how many tiles are contained within the second box, which is 26. - Second I remind myself that 2 selections are being made at random. - Thus, 26 * 25 = 650 then 650/2 = 351. - Therefore, 351 possible combinations will include two tiles. How many possible combinations will include a tile and a ring? - I first note that there are 26 tiles, and 7 rings. - So 26 * 7 = 182 / 2 =91, - Thus, there are 91 possible combinations will include a tile and a ring. How many possible combinations will include at least a tile? - First, how many bags in total? Box one - 15 billiard balls, box two - 26 lettered tiles, box three - 7 rings. So 15 + 26 + 7 = 48 bags in the large box. - So the number of 2 bag combinations is: - (48 * 47) / 2 = 1128 - Of the bags that DO NOT contain tiles, 15 + 7 = 22 bags - (22 * 21) / 2 = 231 combinations that have no tiles - Finally, subtract the combinations of bags that contain 0 tiles from the total cumulative number of bags will give the answer, 897 bag combinations that have at least one tile YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I love this type of problem, so long as I can get the “gist” of it down. Note for future reference that more problems like this will be sent via Question Form, with different number values, etc. for you to critique. Also elaborate further on the last part of that question “How many possible combinations will include at least a tile?” ------------------------------------------------ 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: - For the sake of note-taking, I list that: - - Box 1 = 15 billiard balls - Box 2 = 26 lettered tiles - Box 3 = 7 colored rings - Next I consider that, thinking a few steps ahead, now that there is a prime number being used in place of the normal 2, this is 5 * 4 * 3 * 2 * 1, and NOT /5. I also try to think that, given that some items are listed twice, that their total number of options will decrease as they are chosen. - 15 * 26 * 25 * 7 *14 (note that this lists 15 balls total available for the first choice, and the last signifies a decrease in the available options) - The product of the following will be divided by the number of how many selections are made (whose product will be considered according to a descending fashion) - ( 15 * 26 * 25 * 7 * 14) / (5 * 4* 3 * 2 * 1) ****************************************************************************** - Note that the asterisks above signify either a lapse in judgment or a halt in thinking. I noticed that “ordered” choice was given, so then the usual method of division isn’t necessary, so the solution stands at the product of 15 * 26 * 25 * 7 * 14. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Are there really 955,500 ways this can be achieved? It seems like a hell of a lot for just a box full of 48 items.
.............................................
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): Please provide an ulterior explanation; I seem to be getting bogged down in the cross-referencing between ordered/unordered combinations. - Otherwise I think my original answer stands as the correct one, but I later defined the quotient to be a decimal, which is in this case improbable. - When I compare the statement [15 * 14] / 2 * [26*25]/2 * 7 - I separate the products and the quotients out as follows: - 105 * 325 * 7 = 238,875 But this is the number of ordered choices. Otherwise, there are half as many if they’re unordered, which still yields a decimal answer. - Please critique as 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: - I first note the usage of the word “collection” and compare it to your previous analogy about the “bag” theory in which one assumes a collection is unordered. - 48 bags total, where 26 have tiles and 7 + 15 = 22 bags do not. - 5 bags are chosen, so 22 * 21 * 20 *19 * 18 total ordered choices to contain no tiles. - But, this is a collection and thus unordered, so we consider the usage of the number of selections in descending fashion: 22 * 21 * 20 * 19 * 18 / 5 * 4 * 3 * 2 *1 or 3160080 / 120 = 26334 ways to get a collection of objects that contains no tiles. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I’m beginning to grasp the process a little more each problem, but still let my lack of certainty reflect itself in my confidence ratings. I feel good about them, but not 100%. ------------------------------------------------ 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: - I make a note that, since we’re forming specific combinations of words, that the order in which they are chosen DOES in fact matter in this case, so no dividing by the number of choices made here! - I also make a note that, since each object is replaced (thereby replenishing the number of options for the next sequential choices) the maximum limit of options available is equivalent for each choice. - Next I theorize that: For every ball chosen (3): 15 * 15 * 15 -> For every tile chosen (2) : 15 * 15 * 15 * 26 * 26 = the number of 5-character ‘word’ combinations with 3 numbers and 2 letters are possible. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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'. ********************************************* Question: `q009. Suppose we choose three balls from one through fifteen, without replacing the balls we choose, and place the three chosen balls in order from left to right. We write down the numbers on these balls in order, forming a larger number which can be 3, 4, 5 or 6 digits long (for example if the numbers are 1, 7 and 12 the resulting number would be 1712). Now, in order to get a 3-digit number, the each of the three balls chosen must have a 1-digit number. Thus each of the first three balls must show one of the digits 1 through 9. Some of the following questions are more challenging than others. Give your best thinking on each one: Answer the following in a similar manner: What has to happen in order to end up with a six-digit number? - Each ball chosen (3) must be > 9 or (10, 11, 12, 13, 14, 15). Must contain 2 digits. What has to happen in order to end up with a five-digit number? - 2 balls chosen must be > 9, 1 ball chosen must be < 10. What has to happen in order to end up with a number having two 3's? - Selecting 2 balls whose numbers are 3 or 13. Are there any numbers that could be obtained in more than one way (i.e., by selecting two or more different sets of three balls)? - Possibly. I have to think of a number whose characters relate to either one 2-digit number (> 9) or two 1-digit numbers (<9). - For example, the result would have to be such that I can choose three 1-digit balls. But at the same time, the 2-digit balls must be chosen, so a greater number of digits are required for this to be successful. - 9, 12, 13 91,312. Using one 1-digit number and two 2-digit numbers. This cannot be found in more than one way because we don’t have the number 91, 31, balls in our collection. They only number 1 - 15. 12, 1, 5 1,215. Using one 2-digit number and two 1-digit numbers. This same number can also be found by using two 1-digit numbers and one 2-digit number. 1, 2, and 15. 1,215 I would theorize that, for this to work, I would list the following constraints represented by the previous example: 1) the digits in the resulting number would have to be > 3 2) the numbers chosen for any combinations of the balls must be 1-5 if followed by at least one 2-digit number 3) The numbers chosen for any combinations of the balls MAY be > 5 IF the sequential combinations are 1-digit numbers with at least a 1 following. 4) the second ball chosen would have to be 1 or, if > 1, the next sequential ball chosen must be 1 and must be followed by a number whose ball would create a 2-digit possibility (i.e. 11, 12, 13, 14, 15) It would ultimately depend on whether one would choose two 1-digit combinations and one 2-digit combinations, one 1-digit numbers and two 2-digit numbers, or a mix of the two options. ************************************************************************************* I spotted a possible discrepancy. Since you stated that the balls are to be listed in order, my above theory is thus disproved, since my example of using numbers 12, 1, and 5 aren’t in the ordered specified. Answer the following: How many different three-digit numbers are possible? How many different six-digit numbers are possible? How many possible numbers contain no 1's? How many possible numbers contain four 1's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: How many different three-digit numbers are possible? - In order for any 3-digit numbers to be obtained, the 3 balls chosen must be 1-9, or only 1-digit. - 9 * 8 * 7 = 504 possible 3-digit number combinations. ** Since these are ordered, one would not halve their options as in unordered situations** How many different six-digit numbers are possible? - In order for any 6-digit numbers to be obtained, the 3 balls chosen must be 10-15, or only 2-digits. - 15 * 14 * 13 = 2730 combinations. But, we still have 12, 11, and 10 numbers that are 2-digit in nature. - 12 * 11 * 10 = 1320 combinations. - Therefore I theorize that there are 4050 different 6-digit combinations possible. How many possible numbers contain no 1's? - I consider the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} in relation to Box One. - In order to find out how many numbers contain to 1’s, we have to cut the following. - {1, 10, 11, 12, 13, 14, 15} - So thus the set becomes {2, 3, 4, 5, 6, 7, 8, 9} which contains 8 different balls that do not have a 1 in their number. - 8 * 7 * 6 = 336 different ordered choices. **continued on next page** How many possible numbers contain four 1's? - I consider the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} once more. - In order for four 1’s to appear in 3 choices: 1. At least one of the numbers must be 11 followed by a combination of either 10, 12, 13, 14, or 15. - So this means: - - (11, 10, 12) (11, 12, 10) (10, 11, 12) (12, 11, 10) (10, 12, 11) (12, 10, 11) - (11, 12, 13) (11, 13, 12) (12, 11, 13) (13, 11, 12) (12, 13, 11) (13, 12, 11) - (11, 14, 15) (11, 15, 14) (14, 11, 15) (15, 11, 14) (14, 15, 11) (15, 14, 11) - - There are 18 choices that contain four 1’s. - 18 * 17 * 16 = 4896 combinations that contain four 1’s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: I went into this problem with a reasonable level on confidence, so let my success be praised or my downfall be corrected, and hopefully, comprehension that later follows be praised anyway.