course mth 152 I am so sorry about having messed up the last submit of my work. For some reason it had some of my work from last semester on it and a bunch of other crap. I dont know what happened. I am a little confused about the fundamental counting rule. I understand when I write out the sets but that defeats the purpose of learning the rule. Any help would be appreciated. Thanks, Connie ŻŴJ{lشassignment #001
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11:19:01 `questionNumber 10000 `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 --> (a,b,c) (a,c,b) (b,a,c) (b,c,a) (c,a,b) (c,b,a) abc, acb, bac, bca, cab, cba confidence assessment: 3
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11:19:18 `questionNumber 10000 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 --> self critique assessment: 3
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11:27:48 `questionNumber 10000 `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 --> (a,b,c) abc, acb, bac, bca, cab, cba, aab, aac, aaa, bba, bbc, bbb, cca, ccb, ccc, abb, acc, baa, bcc, caa, cbb, aba, aca, bab, bcb, cac, cbc, confidence assessment: 3
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11:28:12 `questionNumber 10000 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 --> self critique assessment: 3
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11:33:31 `questionNumber 10000 `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, 2 bac, bca 2 ab, ba 2 cab, cba, cb, ca 4 10 confidence assessment: 1
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11:35:46 `questionNumber 10000 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 understood when I was separating but I dont understand the fundamental counting because it asked for 2 letter combos
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11:37:35 `questionNumber 10000 `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, cab, cba I have 6 confidence assessment: 2
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11:39:35 `questionNumber 10000 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 didnt understand you directions. I was making 3 letter words using the first two letters in the set. Now I get it. I just read the directions wrong self critique assessment: 2
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11:44:43 `questionNumber 10000 `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 --> abc, acb, aba, aca, aab, aac, aaa, abb, acc 9 bac, bca, bab, bcb, bba, bbc, bbb, baa, bcc 9 aa, ab, ac, ba, bc, bb 6 cc, ca, cb 3 27 confidence assessment: 3
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11:45:22 `questionNumber 10000 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 --> Am I doing this right? self critique assessment: 1
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11:58:07 `questionNumber 10000 `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 --> abc, abd, acd, adc, adb , acb 6 bac, bad, bca, bcd, bda, bdc, 6 ab, ac, ad, ba, bc, bd 6 da, dc, db, dab, dac, dcb, dca, dbc, dba 9 27 confidence assessment: 2
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12:04:04 `questionNumber 10000 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 --> So I need to look at the set and pick one letter then that leaves three so pick one and that leaves two and pick one and that leaves one so there are 4 possible then there are 3 possible then there are 2 possible then there is 1 possible I am lost on these directions self critique assessment: 2
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12:09:39 `questionNumber 10000 `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, acbd, adbc, abdc, acdb, adcb, bacd, bcad, bdac, bdca, bcda, badc, cabd, cadb, cbad, cbda, cdab, cdba , dabc, dacb, dbca, dbac, dcab, dcba 24 yes confidence assessment: 3
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12:10:17 `questionNumber 10000 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 --> self critique assessment: 3
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12:19:54 `questionNumber 10000 `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 --> You can have 1 paired with a and red 1 b and red 1c red there are 26 letters in the alphabet so 1 can be paired with 26 different numbers and there are 7 colors so 1 can be paired 7 times with a different number There are 15 numbers 15*7*26=2730 confidence assessment: 1
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12:20:30 `questionNumber 10000 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 --> I think I understand this one self critique assessment: 3
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12:22:10 `questionNumber 10000 `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 is 1,3,5,7,9,11,13,15,that are odd numbers that is 8*26*7= 1456 confidence assessment: 3
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12:22:28 `questionNumber 10000 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 --> self critique assessment: 3
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12:24:28 `questionNumber 10000 `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 were 8 odd and there are 5 vowels (a,e,i,o,u) so 8*5*7=280 confidence assessment: 3
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12:24:36 `questionNumber 10000 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 --> self critique assessment: 3
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12:26:04 `questionNumber 10000 `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 21 consonants and 3 colors 7*21*3=441 confidence assessment: 3
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12:26:09 `questionNumber 10000 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 --> self critique assessment: 3
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12:27:44 `questionNumber 10000 `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 --> 7 even 5 vowels and 3 colors 7*5*3=105 confidence assessment: 2
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12:36:34 `questionNumber 10000 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 counting only 3 colors from the preceeding problem not all 7 colors self critique assessment: 2
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12:40:00 `questionNumber 10000 `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 --> If you choose 2 then you may have 1,2 A red there would be 14 number optionsso 14*26*7=2548 confidence assessment: 3
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12:41:38 `questionNumber 10000 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 dont understnd why you use the original 15 when you are taking 2 balls so there are 14 possibilities for 1 self critique assessment: 2
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12:44:32 `questionNumber 10000 `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 --> There are 15 balls so if you choose 3 then it leaves 12, and choose 3 more and it leaves 9 and choose 3 more and it leaves 6 and choose 3 more and it leaves 3 and take those 3 15*12*9*6*3*3=87480 confidence assessment: 2
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12:46:04 `questionNumber 10000 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 --> So I need to just look at the first ball then the second then the third that would give us just 2730 self critique assessment: 2
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