course Mth 152 I submitted this assignment on January 17. I have not seen anything posted on my access page. I am re-sending the work. Am I doing this correctly? Do I run q_a and query for Assignments 1 & 2 and that is everything I submit for the first week's assignments? Thanks, Christina y锬q憺p廜櫷X偕覠反義稅歐恀脑瑦assignment #001
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20:32:34 `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: 2
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20:33:21 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 --> Correct answer. self critique assessment: 3
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20:47:15 `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 --> abc, acb, aaa, abb, acc, aab, aac, aca, aba, bac, bca, bbb, baa, bcc, bba, bbc, bab, bcb, cba, cab, ccc, cbb, caa, ccb, cca, cac, cbc confidence assessment: 2
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20:47:45 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 --> Correct answer. self critique assessment: 3
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21:03:11 `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 --> 3 2 6 1 6 confidence assessment: 1
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21:04:02 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 --> Correct answer! self critique assessment: 3
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21:05:13 `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 Yes, my answer matches my list. confidence assessment: 3
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21:05:43 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 --> Correct answer. self critique assessment: 3
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21:08:51 `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 --> 3 (a, b, or c) 3 (a, b, or c) 27 3 (a, b, or c) 27 choices (3 * 3 * 3 = 27) confidence assessment: 2
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21:09:07 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 --> Correct answer. self critique assessment: 3
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21:15:18 `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 --> 4 (a, b, c, or d) 3 12 24 24 confidence assessment: 1
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21:18:31 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 --> Although I came up with the correct answer of 24 possible three-letter words, I did not solve the problem correctly. I put 24 on the third question because I was thinking of all the possibilities for all letters. self critique assessment: 2
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21:22:19 `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, dacb, dabc, dbac, dbca, dcab, dcba 24 possibilities. Yes, my list confirms my answer to the preceding question. confidence assessment: 2
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21:24:04 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 --> I did four-letter words. A total of 24 possiblities. self critique assessment: 3
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21:27:09 `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 --> To answer this question, I would multiply 15 * 26 * 7 with an answer of 2,730 possibilities. confidence assessment: 2
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21:27:29 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 --> Correct answer. self critique assessment: 3
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21:30:12 `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 numbers between 1 and 15. 8 * 26 * 7 = 1,456 collections contain an odd number. confidence assessment: 1
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21:30:47 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 --> Correct answer. self critique assessment: 3
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21:32:21 `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 --> There are 8 odd numbers between 1 and 15. There are 5 vowels in the alphabet. 8 * 5 * 7 = 280 possible collections contain an odd number and a vowel. confidence assessment: 2
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21:32:39 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 --> Correct answer. self critique assessment: 3
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21:37:30 `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 --> There are 7 even numbers between 1 and 15. There are 21 consonants and three possible colors of the rainbow. 7 * 21 * 3 = 441 possible collections contain an even number, a consonant, and one of the first three colors of the rainbow. confidence assessment: 1
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21:37:46 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 --> Correct answer. self critique assessment: 3
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21:52:05 `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 --> There are 7 even numbers. If an even number is selected, then a consonant could be selected. If an odd number is selected (8 possibilities), then a vowel would have to be selected (5 possibilities). 7 * 21 = 147 8 * 5 = 40 147 + 40 = 187 possible collections confidence assessment: 1
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21:57:53 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 --> I was way off base here. The correct answer should be 1554, not 1555. 1274 + 525 - 245 = 1554. self critique assessment: 2
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22:04:26 `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 --> There would be 210 possibilities for the balls. 15 * 14 = 210 210 * 26 * 7 = 38,220 possible outcomes. confidence assessment: 1
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22:05:21 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 --> I was partially correct. self critique assessment: 2
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22:06:22 `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 = 2,730 possible collections. confidence assessment: 1
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22:07:47 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 --> Again, I was partially correct, but didn't consider the unordered possibility. self critique assessment: 2
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}W儃掳溦{码訇灷芲他金润コ assignment #001 001. Counting Liberal Arts Mathematics II 01-17-2008