#$&* course Mth 152 I need to know how to access the access page. 9/22/14 at 10:18 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: There are 2 'words' that can be formed starting with the first letter, a. They are abc and acb. There are 2 'words' that can be formed starting with the second letter, b. They are bac and bca. There are 2 'words' that can be formed starting with the third letter, c. They are cab and cba. Note that this listing is systematic in that it is alphabetical: abc, acb, bac, bca, cab, cba. When listing things it is usually a good idea to be as systematic as possible, in order to avoid duplications and omissions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: 'q002. List all possible 3-letter 'words' that can be formed from the set of letters { a, b, c } if we allow repetition of letters. Possible 'words' include 'acb' and 'bac' as before; now 'aba' is permitted, as is 'ccc'. Also specify how many words you listed, and how you could have figured out the result without listing all the possibilities. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: acb, abc, aaa, aac, aab, aba, abb, acc, aca bac, bca, bbb, bba, bbc, bab, baa, bcc, bcb ccc, cca, ccb, caa, cab, cac, cba, cbb, cbc confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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? •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: 1. We have two choices for the first letter because there are only two other letters, so if you take one away that leaves you with two. ab, ba 2. We have two choices for the first letter because there are only two letters, just like above, if you take one letter away it would automatically leave you with two options. bc, cb 3. I think we would have six choices that would be able to form the first two letters chosen because you can only have so many different words when you have only 6 letters. ab, ba, ac, ca, bc, cb 4. There would be two choice left for the thrid letter if we have already used the other two letters in the above questions. ca, cb
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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): I am not understanding this question very well. #3, #4, and #5. I will definitely need help understanding this one. ??? ------------------------------------------------ Self-critique Rating: 1
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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): I need help understanding all of this???
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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): ------------------------------------------------ Self-critique Rating: I need help understanding questrion #3.
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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):
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Imagine three boxes: •The first contains a set of billiard balls numbered 1 through 15. •The second contains a set of letter tiles with one tile for each letter of the alphabet. •The third box contains 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 out of 7 for colors of the rainbow. 1 out of 15 with the billard balls 1 out of 26 letters for the alphabet If we were to choose an object from each box, the the equation would be 15*26*7 which would equal 2, 730. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. For the three boxes of the preceding problem, how many of the possible 3-object collections contain an odd number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well if there were 15 boxes then that would equal 8 odd numbers. One, three, five, seven, nine, eleven, thirteen, fifteen. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel? Your solution: If we take the first box of billard balls which is 1 out of 15, then we would have eight possible choices for the collection. In the alphabet box of 1 out of 26, it has 5 possible choices for vowels which are A, E, I, O, and U. In the rainbow color box of 1 out of 7, there are seven choices of odd and vowels in the collection. So, if we take right from the billard balls, 5 from the alphabet, and seven from the rainbow colors, then the equation would look like this; 8*7*5= 280 possible collections that contain an odd number and a vowel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In this case we have 8 possible choices from the first box and, if we consider only a, e, i, o and u to be vowels, we have only 5 possible choices from the second box. We still have 7 possible choices from the third box. •The number of acceptable 3-object collections is now only 8 * 5 * 7 = 280, just a little over 1/10 of the 2730 unrestricted possibilities. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. For the three boxes of the preceding problem, how many of the possible collections contain an even number, a consonant and one of the first three colors of the rainbow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the three boxes, there are 7 even numbers between 1 and 15. I know that because there were 8 odd numners between them. There are 21 consonants in the alphabet since 5 of them are vowels. If we take the first three colors of the rainbow then we have 3 options, so the equation would say; 7*21*3= 441 possible collections that contain and even number, a consonant, and one of the first three colors of the rainbow. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: My self-citique is where I explain everything. 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): I am really confused on how many more vowels and even numbers there are compared to the odd numbers/ vowels.
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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.
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.
If you understand all three of these sentences, then what is the first sentence of the given solution you do not fully understand, and what is your thinking on it?
*@ ------------------------------------------------ Self-critique Rating: STUDENT QUESTION: I’m a little confused about when the order of choice is not important. Because, if you are choosing 3 different objects, and it doesn’t matter what order you chose them in, wouldn’t you still have the same number of possibilities? You would choose one object, and there would be 14 left, and chose the 2nd object, and there would be 13 left for the third choice. I’m sure I’m just not looking at this in the right manner. Is there any way you could explain it a little better? INSTRUCTOR RESPONSE: You would have the same number of ordered possibilities, and you would begin by calculating this number. If the choices are unordered, though, you have to divide this result by the number of ways in which the same objects could be chosen, but in different orders. ********************************************* Question: `q015. How many three-letter 'words' can be formed from the set {a, b, c, d, e}, if letters cannot repeat? In how many of these words will the three letters be in alphabetical order? How many three-letter 'words' are possible if any letter can be used as many times as desired? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I need help with this one as well???Be sure to include the entire document, including my notes. &#
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