course Mth 152 The rest of the assign will follow |a®ԝVLnmԊzassignment #001
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13:51:57 `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, bca, bac, cab, cba 6 words. I made all the words for a then b then c. confidence assessment: 2
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13:53:18 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 --> Listing it systematicly helps me keep it straight. self critique assessment: 2
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13:57:15 `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 --> Instead of writing each letter out, could you mutiply 3*3*3. The 3 letters * 3 letter words * 3 possiblilties. 27 possible words confidence assessment: 2
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13:57:54 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 --> I mutiplied 3*3*3 is that okay to do it like that? self critique assessment: 2
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14:06:06 `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 --> abc, acb, bac, bca, cab, cba 2 for the first letter 2 for the second letter 1 for the first two letters 2 for the third letter 9 for the three letter word confidence assessment: 2
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14:11:02 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 --> I got 9 instead of 6, I think. If I did I mutiplied wrong getting 9 instead of 6. I mutiplied 3*3 and it should have been 3*2 so since there are only 2 sets per letter self critique assessment: 2
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14:11:55 `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 --> abc, acb, bac, bca 2 possible numbers for the first two letters. confidence assessment: 2
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14:13:05 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 --> I forgot to do c, I messed up and thought it was only a & b that I listed. self critique assessment: 2
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14:18:19 `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 --> aaa, aab, aac, aba, aca, abb, acc, acb, abc bbb, bba, bbc, bab, bcb, baa, bcc, bca, bca ccc, cca, ccb, cac, cbc, caa, cbb, cba, ccb First letter 9 Second letter 9 Third letter 9 27 choices for the the 3 letter word confidence assessment: 2
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14:23:45 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 --> I got it this time, I listed them but I would have listed it 3*3*3. self critique assessment: 2
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14:32:31 `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 --> abcd, abdc, acdb, acbd 4 for the first letter 4 for the scond letter 4 for the third letter 24 possiblities confidence assessment: 2
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14:34:48 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 --> I got the total works right but I got each individual group except for the first one wrong. I said that you would get 4 for each group instead of 4, 3, 2 but in my mind I mutiplied 4*3*2 to get the 24. self critique assessment: 2
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