course Math151 Sorry so late. ԸyDzjǙPExOassignment #001
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20:02:39 `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 confidence assessment: 3
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20:07:25 `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 --> tHER EIS 27 POSSIBLES AA: AAB, AAC,AAA AB: ABA, ABB, ABC AC: ACA, ACB, ACC B: BAA, BAB,BBA,BBC,BCB,BCC,BAC,BBB,BCA C: CAA,CAB,CAC,CBA,CBB,CBC,CCA,CCB,CCC confidence assessment: 3
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20:12:07 `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 --> There are three choices in the first letter A, B, C. There are two choices for the second letter. There is one for the third letter. The total possibles is six confidence assessment: 1
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20:17:00 `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 --> Yes ab, ac,ba,bc,ca,cb confidence assessment: 2
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20:20:17 `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 --> there is a total of 27 choices. there is 3 choices in the first letter set 3choices in the second set, and 3 chioces in the third set. 3*3*3=27 confidence assessment: 2
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20:22:57 `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 --> 24 poss. There are four letters for the first letter set, 3 letters for the second set, and 2 letters for the third set. 4*3*2=24 confidence assessment: 2
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20:27:35 `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 --> yes there are 6 possibiltes. abcd, abdc, acbd,acdb,adbc,adab bac,bad,bca,bcd,bda,bdc cab,cad,cba,cbd,cda,cdb dab, dac, dba, dbc,dba,dcb confidence assessment: 1
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20:29:00 `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 balls 26 tiles 7 tiles 15* 26* 7= 2730 poss. confidence assessment: 2
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20:36:02 `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 15 balls 8 labeled with odd #s There are 8 poss. choicesfrom the 1st box 26 in the 2nd box and 7 in the 3rd box. 8*26*7= 1456 confidence assessment: 2
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20:37:51 `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 in box 1, 5 in box 2 and 7 in box 3. 8*5*7=280 confidence assessment: 2
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20:39:35 `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 in the first box, 21 in the second box, and 3 in the last box. 7*21*3=441 poss. collections confidence assessment: 3
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20:44:44 `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 in box 1, 26 in box 2, 7 in box 3. 7*26*7=1274 even numbers 15 in box 1, 5 in box 2, 7 in box 3. 15*5*7=525 vowels 1274+525= 1799 poss even or vowels 7*5*7=245 1274+525-245=1555 confidence assessment: 2
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20:46:34 `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 are 15 ball poss., 14 for the second. There are 26 poss. of tiles 7 for the rings. 15*14*26*7 confidence assessment: 2
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20:49:09 `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 in box 1 14 in box 2, and 13 in box three 3*2*1=6 15*14*13/6 confidence assessment: 2
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D~zGS assignment #002 002. Permutations, combinations, rearranging letters of words. Liberal Arts Mathematics II 02-25-2008
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21:04:00 `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 --> There are 26 poss. for the first tile, 25 poss. for the seconf tile, 24 for the third tile. The three letter words are 26*25*24 confidence assessment: 2
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21:07:54 `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 --> In the 3 tile unordered collection the poss. are 3*2*1=6 26*25*24/6 confidence assessment: 3
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21:10:10 `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 --> The greatest poss. is 14 +15 =29, which is equal but not greater than 29 15*14/2= 105 poss unordered combos confidence assessment: 2
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21:14:23 `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 --> There are 7 rings in bag 1, 6 in bag 2. 7*6/2=21poss. combs There are 26 tiles in bag 1 25 tiles in bag 2 26*25/2=325 poss. combos there are 26 tile and 7 rings, 26*7/2= 91 there are 26 tiles?/ confidence assessment: 2
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21:18:12 `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 --> the choices are listed by the number of objects. there are 15 bags 26 tiles, 25 tiles, 7 rings, 14 different ways 15* 26* 25*7*14 ways confidence assessment: 2
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21:21:06 `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 poss ball out comes 26*25 /2 poss. tile out comes 7 poss. ring choices a equation: [15*14/2]*[26*25/2] *7 confidence assessment: 2
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21:26:17 `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 --> There are 22 bags that donot have tile, when you pick 5 bags. It would then look like this. 22*21*20*19*18 confidence assessment: 3
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21:28:08 `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 --> There are 15 balls, and 26 possible tiles Therefore if have 3 balls and 2 tiles. (the poss. 5 words are) 15*15*15*26*26 confidence assessment: 3
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