#$&* course mth 152 02/03 12 for FEB. 03 around 12:15pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: There are 2 'words' that can be formed starting with the first letter, a. They are abc and acb. There are 2 'words' that can be formed starting with the second letter, b. They are bac and bca. There are 2 'words' that can be formed starting with the third letter, c. They are cab and cba. Note that this listing is systematic in that it is alphabetical: abc, acb, bac, bca, cab, cba. When listing things it is usually a good idea to be as systematic as possible, in order to avoid duplications and omissions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } if we allow repetition of letters. Possible 'words' include 'acb' and 'bac' as before; now 'aba' is permitted, as is 'ccc'. Also specify how many words you listed, and how you could have figured out the result without listing all the possibilities. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cab, cb, bac, bc, ac, A, aba, ccc, abc, bcc,CA I would need to use dictionary to see if there is more word combinations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Listing alphabetically: The first possibility is aaa. The next two possibilities start with aa. They are 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: again listing in alphabetical order we have.baa, bab, bac; bba, bbb, bbc; bca, bcb and bcc 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'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok would need dictionary for more words. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `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? Abc, acb so 2 would be the answer How many choices do we then have for the second letter? Bca, bac So I can count 2 possible chooses. How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? Bc, ab, ba three choses How many choices are then left for the third letter? Cab, cba 2 would be the answer How many choices does this make for the 3-letter 'word'? 6 choices YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: How many choices do we have for the first letter chosen? Abc, acb so 2 would be the answer How many choices do we then have for the second letter? Bca, bac So I can count 2 possible chooses. How many choices do we therefore have for the 2-letter 'word' formed by the first two letters chosen? ab, ac, ba, bc, ca or cb three choses How many choices are then left for the third letter? Cab, cba 2 would be the answer How many choices does this make for the 3-letter 'word'? 6 choices confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 3 choices for the first letter. The choices are a, b and c. Recall that repetition is not permitted. So having chosen the first letter, whichever letter is chosen, there are only 2 possible choices left when we choose the second. The question arises whether there are now 2 + 3 = 5 or 3 * 2 = 6 possibilities for the first two letters chosen. The correct answer is 3 * 2 = 6. This is because for each of the 3 possible choices for the first letter, there are 2 possible choices for the second. [ This result illustrates the Fundamental Counting Principal: If we make a number of distinct choices in a sequence, the total number of possibilities is the product of the numbers of possibilities for each individual choice. ] Returning to the original Self-critique (if necessary): By the time we get to the third letter, we have only one letter left, so there is only one possible choice for our third letter. Thus the first two letters completely determine the third, and there are still only six possibilites. The Fundamental Counting Principal confirms this: the total number of possibilities is the product 3 * 2 * 1 = 6 of the numbers of possibilities for each of the sequential choices. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. Check your answer to the last problem by listing the possibilities for the first two letters. Does your answer to that question match your list? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ab, ac, ba, bc, ca, and cb total of 6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: unlimited choses for you could go on for ever. Allowing repetitions means I can repeat over ABC,ABC there is no end.
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Given Solution: As before there are 3 choices for the first letter. However this time repetition is permitted so there are also 3 choices available 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had different answer ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q006. If we were to form a 3-letter 'word' from the set {a, b, c, d}, without allowing a letter to be repeated, then How many choices would we have for the first letter chosen? Abc just one How many choices would we then have for the second letter? Bad just one How many choices would we therefore have for the 2-letter 'word' formed by the first two letters chosen?ab,ba just two How many choices would then be left for the third letter? Cab one choice How many possibilities does this make for the 3-letter 'word'? 3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Abc just one Bad just one ab,ba just two Cab one choice 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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' which can be formed from the original 4-letter set, provided repetitions are not allowed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): without allowing a letter to be repeatedmeans you can not reuse the letters again so the answer to your problem is only 3
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Given Solution: Listing alphabetically we have abc, abd, acb, acdb, adb, adc; 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. We therefore have a list of 4 * 6 = 24 possible 3-letter words. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): this is your writing below it states one thing but then you give an answer that do not match you wording. without allowing repetition of letters within a word.
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Given Solution: There are 15 possible choices from the first box, 26 from the second, and 7 from the third. By the Fundamental Counting Principle, the total number of possibilities is therefore 15 * 26 * 7 = 2730. It would be possible to list the possibilities. Using the numbers 1, 2, …, 15 for the balls, the lower-case letters a, b, c, …, z for the letter tiles, and the upper-case letters R, O, Y, G, B, I, V for the colors of the rings, the following would be an outline of the list: 1 a R, 1 a O, 1 a Y, ..., 1 a V (seven choices, one for each color starting with ball 1 and the ‘a’ tile) 1 b R, 1 b O, ..., 1 b V, (seven choices, one for each color starting with ball 1 and the ‘b’ tile) 1 c R, 1 c O, ..., 1 c V, (seven choices, one for each color starting with ball 1 and the ‘c’ tile) … … continuing through the rest of the alphabet … 1 z R, 1 z O, …, 1 z V, (seven choices, one for each color starting with ball 1 and the ‘z’ tile) … (this completes all the possible choices with Ball #1; there are 26 * 7 choices, one for each letter-color combination) 2 a R, 2 a O, ..., 2 a V, … 2 z R, 2 z ), …, 2 z V … (consisting of the 26 * 7 possibilities if the ball chosen is #2) etc., etc. 15 a R, 15 a O, ..., 15 a V, … 15 z R, 15 z ), …, 15 z V … (consisting of the 26 * 7 possibilities if the ball chosen is #15) If the complete list is filled out, it should be clear that it will consist of 15 * 26 * 7 possibilities. To actually complete this listing would be possible, not really difficult, but impractical because it would take hours and would be prone to clerical errors. The Fundamental Counting Principle ensures that our result 15 * 26 * 7 is accurate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If content of what is in the boxes is defined then you cannot multiple it. You stated pick one item from the box which is 3 then 48 items are in the boxes in total for the boxes are defined not magical changing to get 15*26*7= ------------------------------------------------ Self-critique Rating:3
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Given Solution: Only the balls are numbered. 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, since no number are represented in either of those boxes. We are unrestricted in our choice 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 unrestricted 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8+13+4= 25 odd choses and 6 vowels y sometimes in engilish= 31 1,3,5,7,9,11,13,15 red, orange, yellow, green, blue, indigo and violet= 4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 the second box. We still have 7 possible choices from the third box. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12,4,6,8,10,12,14=7 consonant 20 if you count y 21 3 possible out of red Orange, yellow 7+21+3=31 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 7 even numbers between 1 and 15, and if we count y as a conontant there are 21 consonants in the alphabet. There are therefore 7 * 21 * 3 = 441 possible 3-object collections containing an even number, a consonant, and one of the first three colors of rainbow. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q012. For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 vowels 26 letter count 2,4,6,8,10,12,14,16,18,20,22,24,26 = 13 15 numbered balls= 2,4,6,8,10,12,14= 7 colored (red, orange, yellow, green, blue, indigo and violet, abbreviated ROY G BIV) =3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. · Therefore 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 union of sets A and B and A^B is their intersection, and n(S) stands for the number of objects in the set S. As the rule is applied here, A is the set of collections containing an even number and B the set of collections containing a vowel, so that A U B is the set of all collections containing a letter or a vowel, and A ^ B is the set of collections containing a vowel and a consonant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q013. For the three boxes of the preceding problems, if we choose two balls from the first box, then a tile from the second and a ring from the third, how many possible collections are there? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2+1+1=4 collection defined : something that is collected; a group of objects or an amount of material accumulated in one location, especially for some purpose or as a result of some process:
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Given Solution: 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 problem as stated is specifies a collection of objects. The word 'collection' specifies how we are to treat the objects. If we were going to place the items in the order chosen, then there would be 15 * 14 * 26 * 7 possibilities. For example, if balls 7 and 12 were chosen, the ordered choice would look different if ball 7 was placed before ball 12 than if they were placed in the reverse order. However a collection is not ordered. For a collection, it's as if we're just going to toss the items into bag with no regard for order, so it doesn't matter which ball is chosen first. Since the two balls in any given collection could have been chosen in either of two orders, there are only half as many possible collections as there are ordered choices. In this case, since the order in which the balls are chosen doesn’t matter, then our answer would that we have just 15 * 14 * 26 * 7 / 2 possible unordered collections. (By contrast, if the order did matter, which is does not for a collection, then our answer would be that there are 15 * 14 * 26 * 7 possible ordered choices.) STUDENT QUESTION I don’t understand what you mean by the /2 at the end of the first part if you say the order chosen doesn’t matter. Why are you dividing by 2? Is that because you are picking two and then the order doesn’t matter at all effectively halving the choices?
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Given Solution: There are 15 possibilities for the first ball chosen, 14 for the second, and 13 for the third. If the choices are going to be placed in the order chosen there are therefore 15 * 14 * 13 possible outcomes. That is, there are 15 * 14 * 13 ordered choices. 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). So if the order of choice is not important, then there are only 1/6 as many possibilities. Thus if the order in which the objects are chosen doesn't matter, there are only 15 * 14 * 13 / 6 possible outcomes. A briefer summary: There are 15 * 14 * 13 ordered choices of three objects. For any three objects, they could appear in 3 * 2 * 1 = 6 possible orders. So the same three objects appear 6 different times among the ordered choices. There are thus only 1/6 as many unordered choices, or collections, as ordered choices. The number of possible collections is therefore 15 * 14 * 13 / 6. STUDENT COMMENT I think I understand how this works sort of but a little bit of clarification on what to do with more than 3 choices (the example given in the solution) would help me out understanding this more clearly. I think I have an idea though.