course Mth 152 NCբ}wzxFSassignment #001
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20:25:53 `q001. Note that there are 14 questions in this assignment. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } without repeating any of the letters. Possible 'words' include 'acb' and 'bac'; however 'aba' is not permitted here because the letter 'a' is used twice.
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RESPONSE --> ABC, ACB, BAC, BCA, CAB, CBA. confidence assessment: 3
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20:26:34 There are 2 'words' that can be formed starting with the first letter, a. They are abc and acb. There are 2 'words' that can be formed starting with the second letter, b. They are bac and bca. There are 2 'words' that can be formed starting with the third letter, c. They are cab and cba. Note that this listing is systematic in that it is alphabetical: abc, acb, bac, bca, cab, cba. It is important when listing things to be as systematic as possible, in order to avoid duplications and omissions.
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RESPONSE --> Ok. self critique assessment: 3
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20:28:59 `q002. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } if we allow repetition of letters. Possible 'words' include 'acb' and 'bac' as before; now 'aba' is permitted, as is 'ccc'.
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RESPONSE --> AAA, AAB, AAC, ABA, ABB, ABC, ACA, ACB, ACC BAA, BAB, BAC, BBA, BBB, BBC, BCA, BCB, BCC CAA, CAB, CAC, CBA, CBB, CBC, CCA, CCB, CCC confidence assessment: 3
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20:29:53 Listing alphabetically the first possibility is aaa. There are 2 more possibilities starting with aa: aab and aac. There are 3 possibilities that start with ab: aba, abb and abc. Then there are 3 more starting with ac: aca, acb and acc. These are the only possible 3-letter 'words' from the set that with a. Thus there are a total of 9 such 'words' starting with a. There are also 9 'words' starting with b: baa, bab, bac; bba, bbb, bbc; bca, bcb and bcc, again listing in alphabetical order. There are finally 9 'words' starting with c: caa, cab, cac; cba, cbb, cbc; cca, ccb, ccc. We see that there are 9 + 9 + 9 = 27 possible 3-letter 'words'.
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RESPONSE --> Ok. self critique assessment: 3
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20:35:43 `q003. If we form a 3-letter 'word' from the set {a, b, c}, not allowing repetitions, then how many choices do we have for the first letter chosen? How many choices do we then have for the second letter? How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices are then left for the third letter? How many choices does this make for the 3-letter 'word'?
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RESPONSE --> A, B, or C could be the first letter. After picking the first letter, only one of the remaining two could be used as a second letter. There are 6 (3*2) choices for the 2-letter word. There's only one choice for the third letter. There are 6 (3*2*1) choices for the 3-letter word. confidence assessment: 3
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20:37:00 There are 3 choices for the first letter. The choices are a, b and c. Recall that repetition is not permitted. So having chosen the first letter, whichever letter is chosen, there are only 2 possible choices left. The question arises whether there are now 2 + 3 = 5 or 3 * 2 = 6 possibilities for the first two letters chosen. The correct choice is 3 * 2 = 6. This is because for each of the 3 possible choices for the first letter, there are 2 possible choices for the second. This result illustrates the Fundamental Counting Principal: If we make a number of distinct choices in a sequence, the net number of possibilities is the product of the numbers of possibilities for each individual choice. By the time we get to the third letter, we have only one letter left, so there is only one possible choice. Thus the first two letters completely determine the third, and there are still only six possibilites. The Fundamental Counting Principal confirms this: the total number possibilities must be 3 * 2 * 1 = 6.
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RESPONSE --> Ok. self critique assessment: 3
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20:38:22 `q004. Check your answer to the last problem by listing the possibilities for the first two letters. Does your answer to that question match your list?
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RESPONSE --> AB, AC, BA, BC, CA, CB. confidence assessment: 3
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20:39:40 Listing helps clarify the situation. The first two letters could be ab, ac, ba, bc, ca or cb. Having determined the first two, the third is determined: for example if the first to letters are ba the third must be c. The possibilities for the three-letter 'words' are thus abc, acb, bac, bca, cab and cba; note that this list is obtained by simply adding the necessary letter to each of the two-letter sequences ab, ac, ba, bc, ca and cb.
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RESPONSE --> Ok. self critique assessment: 3
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20:44:39 `q005. If we form a 3-letter 'word' from the set {a, b, c}, allowing repetitions, then how many choices do we have for the first letter chosen? How many choices do we then have for the second letter? How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices are then left for the third letter? How many choices does this make for the 3-letter 'word'?
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RESPONSE --> 1st letter: 3 possibilities. 2nd letter: 3 possibilities. 3rd letter: 3 possibilities. 3*3*3 = 27 possibilities. confidence assessment: 3
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20:46:04 As before there are 3 choices for the first letter. However this time repetition is permitted so there are also 3 choices for the second letter and 3 choices for the third. By the Fundamental Counting Principal there are therefore 3 * 3 * 3 = 27 possibilities. Note that this result agrees with result obtained earlier by listing.
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RESPONSE --> Ok. self critique assessment: 3
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20:50:07 `q006. If we were to form a 3-letter 'word' from the set {a, b, c, d}, without allowing a letter to be repeated, how many choices would we have for the first letter chosen? How many choices would we then have for the second letter? How many choices would we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices would then be left for the third letter? How many possibilities does this make for the 3-letter 'word'?
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RESPONSE --> 1st letter: 4 possibilities. 2nd letter: 3 possibilities. 2-Letter word: 12 possibilities. 3rd letter: 2 possibilities. 3-Letter word: 24 possibilities. confidence assessment: 3
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20:52:10 The first letter chosen could be any of the 4 letters in the set. The second choice could then be any of the 3 letters that remain. The third choice could then be any of the 2 letters that still remain. By the Fundamental Counting Principal there are thus 4 * 3 * 2 = 24 possible three-letter 'words'.
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RESPONSE --> Ok. self critique assessment: 3
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20:54:01 `q007. List the 4-letter 'words' you can form from the set {a, b, c, d}, without allowing repetition of letters within a word. Does your list confirm your answer to the preceding question?
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RESPONSE --> ABCD, ABDC, ACBD, ACDB, ADBC, ADCB BACD, BADC, BCAD, BCDA, BDAC, BDCA CABD, CADB, CBAD, CBDA, CDAB, CDBA DABC, DACB, DBAC, DBCA, DCAB, DCBA confidence assessment: 3
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20:54:35 Listing alphabetically we have abcd, abdc, acbd, acdb, adbc, adcb; bac, bad, bca, bcd, bda, bdc; cab, cad, cba, cbd, cda, cdb; dab, dac, dba, dbc, dca, dcb. There are six possibilities starting with each of the four letters in the set.
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RESPONSE --> Ok. self critique assessment: 3
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20:57:21 `q008. Imagine three boxes, one containing a set of billiard balls numbered 1 through 15, another containing a set of letter tiles with one tile for each letter of the alphabet, and a third box containing colored rings, one for each color of the rainbow (these colors are red, orange, yellow, green, blue, indigo and violet, abbreviated ROY G BIV). If one object is chosen from each box, how many possibilities are there for the collection of objects chosen?
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RESPONSE --> 15*26*7 = 2730 possibilities confidence assessment: 3
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20:58:49 There are 15 possible choices from the first box, 26 from second, and 7 from the third. The total number of possibilities is therefore 15 * 26 * 7 = 2730. It would be possible to list the possibilities: 1 a R, 1 a O, 1 a Y, ..., 1 a V 1 b R, 1 b O, ..., 1 b V, 1 c R, 1 c O, ..., 1 c V, ... , 1 z R, 1 z O, ..., 1 z V, 2 a R, 2 a O, ..., 2 a V, etc., etc. This listing would be possible, not really difficult, but impractical because it would take hours. The Fundamental Counting Principle ensures that our result is accurate.
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RESPONSE --> Ok. self critique assessment: 3
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21:02:43 `q009. For the three boxes of the preceding problem, how many of the possible 3-object collections contain an odd number?
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RESPONSE --> There are 8 odd balls in the first box, so 8*26*7 = 1456 possibilities. confidence assessment: 3
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21:04:22 The only possible odd number will come from the ball chosen from the first box. Of the 15 balls in the first box, 8 are labeled with odd numbers. There are thus 8 possible choices from the first box which will result in the presence of an odd number. The condition that our 3-object collection include an odd number places no restriction on our second and third choices. We can still choose any of the 26 letters of the alphabet and any of the seven colors of the rainbow. The number of possible collections which include an odd number is therefore 8 * 26 * 7 = 1456. Note that this is a little more than half of the 2730 possibilities. Thus if we chose randomly from each box, we would have a little better than a 50% chance of obtaining a collection which includes an odd number.
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RESPONSE --> Ok. self critique assessment: 3
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21:05:48 `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel?
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RESPONSE --> 8*5*7 = 280 possibilities. confidence assessment: 3
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21:06:46 In this case we have 8 possible choices from the first box and, if we consider only a, e, i, o and u to be vowels, we have only 5 possible choices from second box. We still have 7 possible choices from the third box, but the number of acceptable 3-object collections is now only 8 * 5 * 7 = 280, just a little over 1/10 of the 2730 unrestricted possibilities.
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RESPONSE --> Ok. self critique assessment: 3
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21:08:09 `q011. For the three boxes of the preceding problem, how many of the possible collections contain an even number, a consonant and one of the first three colors of the rainbow?
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RESPONSE --> 7*21*3 = 441 possibilities. confidence assessment: 3
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21:09:46 There are 7 even numbers between 1 and 15, and if we count y as a constant there are 21 consonant in the alphabet. There are therefore 7 * 21 * 3 = 441 possible 3-object collections containing an even number, a consonant, and one of the first three colors of rainbow.
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RESPONSE --> Ok. self critique assessment: 3
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21:10:39 `q012. For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel?
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RESPONSE --> 1274 + 525 = 1799 1799 - (7*5*7 = 245) = 1554 possibilities. confidence assessment: 3
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21:12:02 There are 7 * 26 * 7 = 1274 collections which contain an even number. There are 15 * 5 * 7 = 525 collections which contain a vowel. It would seem that there must therefore be 1274 + 525 = 1799 collections which contain one or the other. However, this is not the case. Some of the 1274 collections containing an even number also contain a vowel, and are therefore included in the 525 collections containing vowels. If we add the 1274 and the 525 we are counting each of these even-number-and-vowel collections twice. We can correct for this error by determining how many of the collections in fact contain an even number AND a vowel. This number is easily found by the Fundamental Counting Principle to be 7 * 5 * 7 = 245. All of these 245 collections would be counted twice if we added 1274 to 525. If we subtract this number from the sum 1274 + 525, we will have the correct number of collections. The number of collections containing an even number or a vowel is therefore 1274 + 525 - 245 = 1555. This is an instance of the formula n(A U B) = n(A) + n(B) - n(A ^ B), where A U B is the intersection of two sets and A^B is their intersection and n(S) stands for the number of objects in the set. Here A U B is the set of all collections containing a letter or a vowel, A and B are the sets of collections containing a vowel and a consonant, respectively and A ^ B is the set of collections containing a vowel and a consonant.
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RESPONSE --> Ok. self critique assessment: 3
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21:14:30 `q013. For the three boxes of the preceding problem, if we choose two balls from the first box, then a tile from the second and a ring from the third, how many possible outcomes are there?
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RESPONSE --> 15*14*26*7 = 38220 possibilities. confidence assessment: 3
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21:14:55 There are 15 possibilities for the first ball chosen, which leaves 14 possibilities for the second. There are 26 possibilities for the tile and 7 for the ring. We thus have 15 * 14 * 26 * 7 possibilities. However the correct answer really depends on what we're going to do with the objects. This has not been specified in the problem. For example, if we are going to place the items in the order chosen, then there are indeed 15 * 14 * 26 * 7 possibilities. On the other hand, if we're just going to toss the items into a box with no regard for order, then it doesn't matter which ball was chosen first. Since the two balls in any collection could have been chosen in either of two orders, there are only half as many possibilities: we would have just 15 * 14 * 26 * 7 / 2 possible ways to choose an unordered collection.
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RESPONSE --> Ok. self critique assessment: 3
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21:16:11 `q014. For the three boxes of the preceding problem, if we choose only from the first box, and choose three balls, how many possible collections are there?
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RESPONSE --> 15*14*13 = 2730 possibilities. confidence assessment: 3
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21:16:32 There are 15 possibilities for the first ball chosen, 14 for the second, and 13 for the third. If the collection is going to be placed in the order chosen there are therefore 15 * 14 * 13 possible outcomes. On the other hand, if the collections are going to be just tossed into a container with no regard for order, then there are fewer possible outcomes. Whatever three objects are chosen, they could have been chosen in any of 3 * 2 * 1 = 6 possible orders (there are 3 choices for the first of the three objects that got chosen, 2 choices for the second and only 1 choice of the third). This would mean that there are only 1/6 has many possibilities. So if the order in which the objects are chosen doesn't matter, there are only 15 * 14 * 13 / 6 possibilities.
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RESPONSE --> Ok. self critique assessment: 3
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ߛ딂\̊͂ assignment #002 002. Permutations, combinations, rearranging letters of words. Liberal Arts Mathematics II 06-02-2007
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21:37:48 `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 --> First tile: 26 possibilities Second tile: 25 possibilities Third tile: 24 possibilities 26*25*24 = 15600 possibilities. confidence assessment: 3
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21:40:17 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 --> Ok. self critique assessment: 3
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21:41:59 `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*25*24 = 15600/6 = 2600 possibilities. confidence assessment: 3
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21:42:36 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 --> Ok. self critique assessment: 3
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21:46:38 `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 / 2 = 105 possiblities. Since 29 (15+14) is the highest combination, there are 104 possible outcomes less than 29. confidence assessment: 3
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21:47:11 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 --> Ok. self critique assessment: 3
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22:00:16 `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 --> 7*6 / 2 = 21 combinations with 2 rings. 26*25 / 2 = 325 combinations with 2 tiles. 26*7 / 2 = 91 combinations with 1 tile and 1 ring. 48*47 / 2 = 1128 - (22*21 / 2 = 231) = 897 combinations with a tile. confidence assessment: 3
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22:01:19 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. self critique assessment: 3
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22:04:58 `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 --> Ball first: 15 possibilities Tile second: 26 possibilities Tile third: 25 possibilities Ring fourth: 7 possibilities Ball fifth: 14 possibilities 15*26*25*7*14 = 955500 possibilities confidence assessment: 3
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22:07:10 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 --> Ok. self critique assessment: 3
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22:09:25 `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*14 / 2) * (26*25 / 2) * 7 = 238875 possibilities confidence assessment: 3
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22:10:39 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 --> Ok. self critique assessment: 3
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22:13:07 `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 --> 22*21*20*19*18 = 3160080 possibilities. confidence assessment: 3
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22:13:25 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 --> Ok. self critique assessment: 3
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22:16:41 `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*15*15*26*26 = 2281500 possibilities. confidence assessment: 3
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22:17:00 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 --> Ok. self critique assessment: 3
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¾Ъ[gxۺMӑ assignment #003 003. C(n,r) and P(n,r) Liberal Arts Mathematics II 06-02-2007
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22:23:37 `q001. Note that there are 13 questions in this assignment. As we have seen if we choose, say, 3 objects out of 10 distinct objects the number of possible results depends on whether order matters or not. For the present example if order does matter there are 10 choices for the first selection, 9 for the second and 8 for the third, giving us 10 * 9 * 8 possibilities. However if order does not matter then whatever three objects are chosen, they could have been chosen in 3 * 2 * 1 = 6 different orders. This results in only 1/6 as many possibilities, or 10 * 9 * 8 / 6 possible outcomes. We usually write this number as 10 * 9 * 8 / (3 * 2 * 1) in order to remind us that there are 10 * 9 * 8 ordered outcomes, but 3 * 2 * 1 orders in which any three objects can be chosen. If we were to choose 4 objects out of 12, how many possibile outcomes would there be if the objects were chosen in order? How many possible outcomes would there be if the order of the objects did not matter?
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RESPONSE --> Order matters: 12*11*10*9 = 11880 possibilities. Oder doesn't matter: 12*11*10*9 / (4*3*2*1 = 24) = 495 possibilities. confidence assessment: 3
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22:25:38 When choosing 4 objects out of 12, there are 12 choices for the first, 11 choices for the second, 10 choices for the third and 9 choices for the fourth object. If the order matters there are therefore 12 * 11 * 10 * 9 possible outcomes. If the order doesn't matter, then we have to ask in how many different orders any given collection of 4 objects could be chosen. Given any 4 objects, there are 4 choices for the first, 3 choices for the second, 2 choices for the third and 1 choice for the fourth. There are thus 4 * 3 * 2 * 1 orders in which a given set of 4 objects could be chosen. We therefore have 12 * 11 * 10 * 9 / ( 4 * 3 * 2 * 1) possible outcomes when order doesn't matter.
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RESPONSE --> Ok. self critique assessment: 3
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22:28:59 `q002. If order does not matter, then how many ways are there to choose 5 members of a team from 23 potential players?
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RESPONSE --> 22*21*20*19 / (5*4*3*2*1 = 120) = 1463 possibilities. confidence assessment: 3
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22:29:12 If order did matter then there would be 23 * 22 * 21 * 20 * 19 ways choose the five members. However order does not matter, so we must divide this number by the 5 * 4 * 3 * 2 * 1 ways in which any given set of five individuals can be chosen. We therefore have 23 * 22 * 21 * 20 * 19 / ( 5 * 4 * 3 * 2 * 1) possible 5-member teams.
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RESPONSE --> Ok. self critique assessment: 3
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22:30:20 `q003. In how many ways can we line up 5 different books on a shelf?
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RESPONSE --> 5*4*3*2*1 = 120 possibilities. confidence assessment: 3
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22:30:32 It should be clear that there are 5 * 4 * 3 * 2 * 1 ways, since there are 5 choices for the first book, 4 for the second, etc.. If we multiply these numbers out we get 5 * 4 * 3 * 2 * 1 = 120. It might be a little bit surprising that there should be 120 ways to order only 5 objects.
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RESPONSE --> Ok. self critique assessment: 3
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22:32:55 `q004. The expression 5 * 4 * 3 * 2 * 1 is often written as 5 ! , read 'five factorial'. More generally if n stands for any number, then n ! stands for the number of ways in which n distinct objects could be lined up. Find 6 ! , 7 ! and 10 ! .
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RESPONSE --> 6! = 6*5*4*3*2*1 = 720 7! = 7*6*5*4*3*2*1 = 5040 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800 confidence assessment: 3
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22:33:15 6 ! = 6 * 5 * 4 * 3 * 2 * 1 = 720. 7 ! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 10 ! = 3,628,800. These numbers grow at an astonishing rate. The last result here shows is that there are over 3 million ways to arrange 10 people in a line. The rapid growth of these results like in part explain the use of the ! symbol to designate factorials.
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RESPONSE --> Ok. self critique assessment: 3
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22:35:30 `q005. What do we get if we simplify the expression (10 ! / 6 !) ?
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RESPONSE --> 10*9*8*7* (6*5*4*3*2*1) / (6*5*4*3*2*1) = 10*9*8*7 confidence assessment: 2
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22:36:06 10 ! / 6 ! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 6 * 5 * 4 * 3 * 2 * 1). We can simplify this by rewriting it as 10 * 9 * 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1) / ( 6 * 5 * 4 * 3 * 2 * 1) = 10 * 9 * 8 * 7. We see that the 6 * 5 * 4 * 3 * 2 * 1 in the numerator matches the same expression in the denominator, so when divided these expressions give us 1 and we end up with just 10 * 9 * 8 * 7 * 1 = 10 * 9 * 8 * 7. Note that this is just the number of ways in which 4 objects can be chosen, in order, from a collection of 10 objects.
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RESPONSE --> Ok. self critique assessment: 3
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22:37:38 `q006. We saw above that there are 23 * 22 * 21 * 20 * 19 ways to choose 5 individuals, in order, from 23 potential members. How could we express this number as a quotient of two factorials?
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RESPONSE --> 23*22*21*20*19 = 23! / 18! confidence assessment: 3
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22:38:32 If we divide 23 ! by 18 ! , the numbers from 18 down to 1 will occur in both the numerator and denominator and when we divide we will be left with just the numbers from 23 down to 19. Thus 23 * 22 * 21 * 20 * 19 = 23 ! / 18 !.
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RESPONSE --> Ok. self critique assessment: 3
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22:39:52 `q007. How could we express the number of ways to rank 20 individuals, in order, from among 100 candidates?
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RESPONSE --> 100*99*98*97*96.....*81 = 100! / 80! confidence assessment: 1
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22:41:22 There are 100 choices for the first candidate, 99 for the second, 98 for the third, etc.. For the 20th candidate there are 81 choices. You should convince yourself of this if you didn't see it originally. Our product is therefore 100 * 99 * 98 * ... * 81, which can be expressed as 100 ! / 80 !. We see that the denominator must be (100 - 20)! . For this example 100 - 20 represents the difference between the number of individuals available and the number selected.
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RESPONSE --> Ok. self critique assessment: 3
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22:42:50 `q008. How could we express the number of ways to rank r individuals from a collection of n candidates?
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RESPONSE --> N! / (N-R)! confidence assessment: 3
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22:44:25 By analogy with the preceding example, we should divide n ! by ( n - r ) !. The number is therefore n ! / ( n - r ) !.
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RESPONSE --> Ok. self critique assessment: 3
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22:50:37 `q009. The expression n ! / ( n - r ) ! denotes the number of ways in which r objects can be chosen, in order, from among n objects. When we choose objects in order we say that we are 'permuting' the objects. The expression n ! / ( n - r ) ! is therefore said to the the number of permutations of r objects chosen from n possible objects. We use the notation P ( n , r ) to denote this number. Thus P(n, r) = n ! / ( n - r ) ! . Find P ( 8, 3) and explain what this number means.
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RESPONSE --> P(8, 3) is the amount of ways 3 objects within 8 objects can be chosen in order. P(8, 3) = 8! / (8-3)! = 8! / 5! = 8*7*6*5*4*3*2*1 / (5*4*3*2*1) = 8*7*6 = 336 possibilities. confidence assessment: 3
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22:51:17 P(n, r) = n! / ( n - r) !. To calculate P(8, 3) we let n = 8 and r = 3. We get P(8, 3) = 8 ! / ( 8 - 3) ! = 8 ! / 5 ! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / ( 5 * 4 * 3 * 2 * 1) = 8 * 7 * 6. This number represents the number of ways 3 objects can be chosen, in order, from 8 objects.
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RESPONSE --> Ok. self critique assessment: 3
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22:53:41 `q010. In how many ways can an unordered collection of 3 objects be chosen from 8 candidates?
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RESPONSE --> 8*7*6 / (3*2*1) = 56 possibilities. confidence assessment: 3
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22:54:01 There are 8 * 7 * 6 ways to choose 3 objects from 8, in order, and 3! ways to order any unordered collection of 3 objects, so there are 8 * 7 * 6 / ( 3 * 2 * 1 ) possible unordered collections. This number is easily enough calculated. Since 3 goes into 6 twice and 2 goes into 8 four times, we see that 8 * 7 * 6 / ( 3 * 2 * 1) = 4 * 7 * 2 = 56. There are 56 different unordered collections of 3 objects chosen from 8.
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RESPONSE --> Ok. self critique assessment: 3
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22:54:54 `q010. How could the result of the preceding problem be expressed purely in terms of factorials?
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RESPONSE --> 8! / (5! * 3!) confidence assessment: 3
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22:55:04 The product 8 * 7 * 6 is just 8 ! / 5 !, and the expression 8 * 7 * 6 / ( 3 * 2 * 1) can therefore be expressed as 8 ! / ( 5 ! * 3 !).
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RESPONSE --> Ok. self critique assessment: 3
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22:57:45 `q011. In terms of factorials, how would we express the number of possible unordered collections of 5 objects chosen from 16?
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RESPONSE --> 16! / [ (16-5)! * 5!) ] confidence assessment: 3
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22:58:57 There are 16 ! / ( 16 - 5) ! Possible ordered sets of 5 objects chosen from the 16. There are 5 ! ways to order any unordered collection of 5 objects. There are thus 16 ! / [ ( 16 - 5 ) ! * 5 ! ] possible unordered collections of 5 objects from the 16.
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RESPONSE --> Ok. self critique assessment: 3
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23:00:41 `q012. In terms of factorials, how would we express the number of possible unordered collections of r objects chosen from n objects?
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RESPONSE --> P(N,R) / R! = N! / [ R! * (N-R)! ] confidence assessment: 2
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23:03:43 There are P(n, r) = n ! / ( n - r ) ! possible ordered collections of r objects. There are r ! ways to order any unordered collection of r objects. There are thus P ( n, r ) / r! = n ! / [ r ! * ( n - r) ! ] possible unordered collections of r objects chosen from n objects.
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RESPONSE --> Ok. self critique assessment: 3
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23:04:31 `q013. When we choose objects without regard to order, we say that we are forming combinations as opposed to permutations, which occur when order matters. The expression we obtained in the preceding problem gives us a formula for combinations: C ( n , r ) = P ( n, r) / r! = n ! / [ r ! ( n - r) ! ] gives us to number of possible combinations, or unordered collections, of r objects chosen from a set of n objects.
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RESPONSE --> Ok. confidence assessment: 3
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|a}y assignment #001 001. `query 1 Liberal Arts Mathematics II 06-03-2007
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01:34:13 query 11.1.6 {Andy, Bill, Kathy, David, Evelyn}. In how many ways can a secretary, president and treasuer be selected if the secretary must be female and the others male?
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RESPONSE --> Kathy, Andy, Bill; Kathy, Bill, Andy; Kathy, David, Andy; Kathy, Andy, David; Kathy, David, Bill; Kathy, Bill, David. Evelyn, Andy, Bill; Evelyn, Bill, Andy; Evelyn, David, Andy; Evelyn, Andy, David; Evelyn, David, Bill; Evelyn, Bill, David.
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01:34:39 ** Using letters for the names, there are 12 possibilities: kab, kba, kdb, kbd, kda, kad, edb, ebd, eba, eab, eda, ead. There are two women, so two possibilities for the first person selected. The other two will be selected from among the three men, so there are 3 possibilities for the second person chosen, leaving 2 possibilities for the third. The number of possiblities is therefore 2 * 3 * 2 = 12. **
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RESPONSE --> Ok.
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01:36:32 query 11.1.12,18 In how many ways can the total of two dice equal 5?
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RESPONSE --> 1+4 2+3 3+2 4+1
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01:36:59 ** Listing possibilities on first then second die you can get 1,4, or 2,3 or 3,2 or 4,1. There are Four ways. **
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RESPONSE --> Ok.
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01:37:49 In how many ways can the total of two dice equal 11?
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RESPONSE --> 5+6 6+5
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01:38:02 ** STUDENT SOLUTION AND INSTRUCTOR RESPONSE: There is only 1 way the two dice can equal 11 and that is if one lands on 5 and the other on 6 INSTRUCTOR RESPONSE: There's a first die and a second. You could imagine that they are painted different colors to distinguish them. You can get 5 on the first and 6 on the second, or vice versa. So there are two ways. **
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RESPONSE --> Ok.
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01:43:16 query 11.1.36 5-pointed star, number of complete triangles How many complete triangles are there in the star and how did you arrive at this number?
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RESPONSE --> 10 triangles
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01:44:22 ** If you look at the figure you see that it forms a pentagon in the middle (if you are standing at the very center you would be within this pentagon). Each side of the pentagon is the side of a unique triangle; the five triangles formed in this way are the 'spikes' of the star. Each side of the pentagon is also part of a longer segment running from one point of the start to another. This longer segment is part of a larger triangle whose vertices are the two points of the star and the vertex of the pentagon which lies opposite this side of the pentagon. There are no other triangles, so we have 5 + 5 = 10 triangles. *&*&, BDE and CDE. Each of these is a possible triangle, but not all of these necessarily form triangles, and even if they all do not all the triangles will be part of the star. You count the number which do form triangles and for which the triangles are in fact part of the star. **
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RESPONSE --> Ok.
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01:49:36 query 11.1.40 4 x 4 grid of squares, how many squares in the figure?
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RESPONSE --> 30 squares
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01:50:27 ** I think there would be 16 small 1 x 1 squares, then 9 larger 2 x 2 squares (each would be made up of four of the small squares), 4 even larger 3 x 3 squares (each made up of nin small squares) and one 4 x 4 square (comprising the whole grid), for a total of 30 squares. Do you agree? **
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RESPONSE --> Ok; I agree.
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01:51:38 query 11.1.50 In how many ways can 30 be written as sum of two primes?
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RESPONSE --> 19+11 23+7 17+13
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01:52:08 **STUDENT SOLTION AND INSTRUCTOR COMMENT: There are 4 ways 30 can be written as the sum of two prime numbers: 29 + 1 19 + 11 23 + 7 17 + 13 INSTRUCTOR COMMENT: Good, but 1 isn't a prime number. It only has one divisor. **
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RESPONSE --> Ok.
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01:57:42 query 11.1.60 four adjacent switches; how many settings if no two adj can be off and no two adj can be on
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RESPONSE --> Off, On, Off, On On, Off, On, Off
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02:01:08 ** There are a total of 16 settings but only two have the given property of alternating off and on. If the first switch is off then the second is on so the third is off so the fourth is on. If the first is off then then the second is on and the third is off so the fourth is on. So the two possibilies are off-on-off-on and on-off-on-off. If we use 0's and 1's to represent these possibilities they are written 0101 and 1010. **
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RESPONSE --> Ok.
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02:04:33 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None
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02:06:26 ** STUDENT COMMENT: No suprises and it's early so i'm reaching for insight as a child reaches for a warm bottle of milk I would like the answers to all the problems I worked in Assignment 11.1. I was surprised that you only ask for a few. I could not answer 11.1. 63 - What is a Cartesain plane? I could not find it in the text. INSTRUCTOR RESPONSE: I ask for selected answers so you can submit work quickly and efficiently. I don't provide answers to all questions, since the text provides answers to most of the odd-numbered questions. Between those answers and and comments provided here, most people get enough feedback to be confident in the rest of their work. Also I don't want people to get in the habit of 'working backward' from the answer to the solution. If you want to send in your work on other problems, including a full descripton of your reasoning, I'm always glad to look at them. You would have to make those problems self-contained (tell me enough about the problem so I know what the problem is), since I don't always respond from the place where I have my copy of the text. The Cartesian Plane is a plane defined by an x axis and a y axis, on which you can specify points by their coordinates. **
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RESPONSE --> Ok.
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Jگ௳ܛԣÈyؘON~ assignment #002 002. `query 2 Liberal Arts Mathematics II 06-03-2007
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02:13:54 query 11.2.12 find 10! / [ 4! (10-4)! ] without calculator
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RESPONSE --> 10*9*8*7*6*5*4*3*2*1 / [(4*3*2*1) * (6*5*4*3*2*1)] 10*9*8*7 / (4*3*2*1) 5*3*2*7 = 210
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02:14:30 ** 10! / [ 4! * (10-4) ! ] can be simplified to get 10! / ( 4! * 6! ). This gives you 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] . The numerator and denominator could be multiplied out but it's easier and more instructive to divide out like terms. Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1). Every factor of the denominator divides into the numerator without remainder: Divide 4 into 8, divide 3 into 9 and 2 into 10 and you get 5 * 3 * 2 * 7 = 210. NOTE ON WHAT NOT TO DO: You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divided 3628800 by 16480. But that would process would lose accuracy and be ridiculously long for something like 100 ! / ( 30! * 70!). Much better to simply divide out like factors until the denominator goes away. **
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RESPONSE --> Ok.
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02:17:00 query 11.2.25 3 switches in a row; fund count prin to find # of possible settings
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RESPONSE --> 2*2*2 = 8
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02:17:56 ** There are two possible settings for the first switch, two for the second, two for the third. The setting of one switch in independent of the setting of any other switch so the fundamental counting principle holds. There are therefore 2 * 2 * 2 = 8 possible setting for the three switches. COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways. INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. **
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RESPONSE --> Ok.
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02:19:00 query 11.2.27 If no two adjacent switches are off why does the fundamental counting principle not apply?
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RESPONSE --> The counting principle needs the events to be independent of each other.
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02:19:20 ** The reason is that the Fund. Counting Principle requires that the events be independent. Here we have the state of one switch influencing the state of its neighbors (neither neighbor can be the same as that switch). The Fund. Counting Principle requires that the events be independent. **
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RESPONSE --> Ok.
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02:21:59 query 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}?
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RESPONSE --> 3*3*2 = 18
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02:22:40 ** Using the box method: 1st can be any of the three so the first number of possibilities is 3 2nd number can also be any of the three so the second number of possibilities is 3 The last digit must be odd, so there are only 2 choices for it. We therefore have 3*3*2=18 possible combinations.
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RESPONSE --> Ok.
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02:23:35 query 11.2.50 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups
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RESPONSE --> 10*4*6*3 = 720 possibilities
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02:24:01 ** A setup consists of a guitar, a case, an amp and a processor. There are 10 choices for the guitar, 4 for the case, 6 for the amp and 3 for the processor. So there are 10 * 4 * 6 * 3 = 720 possible setups. **
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RESPONSE --> Ok.
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02:24:44 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None.
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xߩEҦ assignment #003 003. `query 3 Liberal Arts Mathematics II 06-03-2007
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02:32:56 Query 11.3.20 5 prizes among 25 students
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RESPONSE --> 25*24*23*22*21 = 6375600 possibilities
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02:33:50 ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 This is P(25, 5). We use permutations because order matters when there are 5 different prizes. **
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RESPONSE --> Ok.
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02:34:27 Is repetition allowed in this situation?
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RESPONSE --> No, because there are only 5 prizes.
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02:34:37 ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people **
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RESPONSE --> Ok.
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02:35:43 Query 11.3.30 3-letter monogram all letters different
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RESPONSE --> 26*25*24 = 600 possibilities
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02:36:58 ** You are choosing 3 different letters, and since the monogram will be different if you change the order we can say that order definitely applies. If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices. We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first). So there are only 25 * 24 = 600 possibilities. **
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RESPONSE --> Ok.
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02:40:58 Query 11.3.42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped?
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RESPONSE --> C(25,3)*C(22,4)*C(18,5)*C(13,6)*C(7,7)
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02:42:03 ** If we make the group of 3 first there are C(25, 3) possible choices. If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices. If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices. If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices. If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices. The Fundamental Counting Principle tells you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group: C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7). Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do this. If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified can get them both into the same form 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ], which can then be further simplified by cancellation and then multiplied. **
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RESPONSE --> Ok.
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02:48:03 Query 11.3.60 C(n,0)What is the value of C(n,0)?What is the value of C(8,0)?
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RESPONSE --> C(N,0) is nothing. C(8,0) is 1.
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02:49:56 ** This is equivalent to choosing 0 objects from n objects. No matter what n is, there is one way to do this, which is to choose nothing. As another example there are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) tells you how many ways that are to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us this: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. **
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RESPONSE --> Ok.
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