#$&* course MTH 152 Question: `q001. Note that there are 16 questions in this assignment.
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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: ********************************************* 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: For this we need to examine each letter individually starting with the letter A. The list of words that begin with the letter A and allowing repeats are as follows: aaa, aab, aac, abb, aba, abc, acc, aca, acb which equals 9 words. Which leaves us with a total of 9 words that start with the letter A repeating. Next we need to look at the letter B. bbb, baa, bcc, bab, bac, bba, bbc, bca, bcb Thus leaving us with a total of 9 words as well. Next we need to look at the letter C ccc, cbb, caa, cab, cac, cbc, cba, ccb, cca Thus showing us 9 words can be made. Now in order to specify how many words we have obtained we simply add 9 letter A words to 9 letter B words to 9 letter C words which gives us 27 total answers. In order to wright this as a mathematical equation we simply list it as follows: 9 + 9 + 9 = 27 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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): I should have listed words that I found with a bold font, also I could have kept it short at the end with just the mathematical equation. ------------------------------------------------ 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: Question #1: We know just by looking at it that we have a total of 3 letters. Question #2: Looking at this we would automatically assume 3 however that is not correct because each letter contains a combination of the other two so the answer should be 2. Question #3: The way we would find the answer to this problem is by simply saying we have 3 letters with 2 possible choices per letter. Thus the equation we would use is 3 letter multiplied by 2 possible choices which would give us 6 answers. The mathematical equation should look like the following: 3 * 2 = 6 Question #4: We know that if the first two letters are chosen then we only have 1 possible choice for the last letter. Question #5: To find out how many word choices these can make all we have to do is apply what we already know from the previous questions and plug them in as a multiplication equation. This would be as follows: Number of letter multiplied by the number of possible choices multiplied by the number of choices we have once the first two letters are chosen. Equals the amount of words we can make with this combination. Thus 3 * 2 * 1 = 6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are 3 choices for the first letter. The choices are a, b and c. Recall that repetition is not permitted. So having chosen thefirst letter, whichever letter is chosen, there are only 2 possible choices left when we choose the second. The question arises whether there are now 2 + 3 = 5 or 3 * 2 = 6 possibilities for the first two letters chosen. The correct answer is 3 * 2 = 6. This is because for eachof the 3 possible choices for the first letter, there are 2possible choices for the second. [ This result illustrates the Fundamental Counting Principal: If we make a number of distinct choices in a sequence, the total number of possibilities is the product of the numbers ofpossibilities for each individual choice. ] Returning to the original Self-critique (if necessary): By the time we get to the third letter, we have only one letter left, so there is only one possible choice for our third letter. Thus the first two letters completely determine the third, and there are still only six possibilites. The Fundamental Counting Principal confirms this: thetotal number of possibilities is the product 3 * 2 * 1 = 6 of the numbers of possibilities for each of the sequentialchoices. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Check your answer to the last problem by listing the possibilities for the first two letters. Does your answer to that question match your list? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The possibilities are as follows: ab, ac, ba, bc, ca, and cb. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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: ********************************************* 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: For the first letter we have 3 choices For the second letter we also have 3 choices as repetitions are allowed For the third question in order to find this information out all we must do is multiply as before 3 * 3 = 9 so the answer is 9 possible choices. For the fourth question since repetition is allowed we know that the answer is 3 For the fifth question all we have to do to find out this information is simply multiply like so; 3 choices multiplied by 3 choices of the second letter multiplied by 3 letters equals a total of 27 combinations. Thus the equation would be 3 * 3 * 3 = 27 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: As before there are 3 choices for the first letter. However this time repetition is permitted so there are also 3 choices available for the second letter and 3 choices for thethird. By the Fundamental Counting Principal there are therefore 3 * 3 * 3 = 27 possibilities. Note that this result agrees with result obtained earlier bylisting. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. If we were to form a 3-letter 'word' from the set {a, b, c, d}, without allowing a letter to be repeated, then How many choices would we have for the first letter chosen? How many choices would we then have for the second letter? How many choices would we therefore have for the 2-letter 'word' formed by the first two letters chosen? How many choices would then be left for the third letter? How many possibilities does this make for the 3-letter 'word'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We would have four choices We would have three choices We would have two choices To find this answer we need to place everything we have into an equation which should look like the following: 4 * 3 * 2 = 24 Thus we have 24 possibilities for forming a 3 letter word from the four letters given. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 thatremain. 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 theoriginal 4-letter set, provided repetitions are not allowed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. List the 3-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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: abc, abd, acb, acdb, adb, adc, bac, bad, bca, bcd, bda, bdc, cab, cad, cba, cbd, cda, cdb, dab, dac, dba, dbc, dca, dcb We can see from these words that this answer does in fact confirm the answer in the previous question because each letter has six different words so we end up multiplying 4 letters * 6 words each = 24 possibilities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Listing alphabetically we have abc, abd, acb, acdb, adb, adc; bac, bad, bca, bcd, bda, bdc; cab, cad, cba, cbd, cda, cdb; dab, dac, dba, dbc, dca, dcb. There are six possibilities starting with each of the four letters in the set. We therefore have a list of 4 * 6 = 24 possible 3-letter words. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Imagine three boxes: The first contains a set of billiard balls numbered 1 through 15. The second contains a set of letter tiles with one tile foreach letter of the alphabet. The third box contains colored rings, one for each colorof the rainbow (these colors are red, orange, yellow, green, blue, indigo and violet, abbreviated ROY G BIV). If one object is chosen from each box, how many possibilitiesare there for the collection of objects chosen? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the first box we know that we have 15 possibilities. From the second box we know there are 26 letters in the alphabet so knowing this we would have 26 possible choices from the second box. From the third box we are given 7 colors which gives us 7 different possible choices. By knowing all of the information above, we would simply plug all this information into a multiplication equation. Which would be as follows: 15 * 26 * 7 = 2730, so now the answer should be 2730 different possibilities. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are 15 possible choices from the first box, 26 from thesecond, 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 foreach 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 hoursand would be prone to clerical errors. The Fundamental Counting Principle ensures that our result 15 * 26 * 7 is accurate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After seeing the answer I should have used the ROY G BIV letters instead of just using the colors. ------------------------------------------------ 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: Going from the answer above only 8 balls have odd numbers. So knowing this we know that there are only 8 possibilities which will give us an odd number. We know that we only need to multiply from here using 8 balls, 26 letters, and 7 colors. So we should have the equation: 8 * 26 * 7 = 1456 Then if we divide our original answer above by 1456 we get the following solution: 2730 / 1456 = 1.875 which tells us that the answer we get is just above half. Thus giving us a 50% chance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Only the balls are numbered. Of the 15 balls in the first box, 8 are labeled with odd numbers. There are thus 8 possible choices from the first box which will result in the presence of an odd number. The condition that our 3-object collection include an odd numberplaces no restriction on our second and third choices, since no number are represented in either of those boxes. We areunrestricted in our choice any of the 26 letters of the alphabet and any of the seven colors of the rainbow. The number of possible collections which include an odd number is therefore 8 * 26 * 7 = 1456. Note that this is a little more than half of the 2730 unrestricted possibilities. Thus if we chose randomly from each box, we would have a little better than a 50% chance of obtaining a collectionwhich includes an odd number. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I needed to explain things in a better fashion than what I provided. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 8 possible choices in the first box. Since the vowels are A, E, I, O, U then this means we have 5 possible choices in the second box. According to the question we will not add or take anything away from the amount of colors there for leaving a total of 7 possible choices in the third box. The amount of possible collections is as follows: 8 * 5 * 7 = 280 Thus the answer is 280 possible collections. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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 only8 * 5 * 7 = 280, just a little over 1/10 of the 2730 unrestricted possibilities. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I needed to take it one step farther and show the divishion part of this. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. For the three boxes of the preceding problem, how many of the possible collections contain an even number, aconsonant and one of the first three colors of the rainbow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know there are 7 even numbers between 1 and 15 and if we use Y as a consonants we know just from looking at it that there are 21 consonants in the alphabet. So now we know that in order to find the answer we take the amount of even numbers multiplied by the about of consonants multiplied by how many are in each combination, and we get the following answer. 7 * 21 * 3 = 441 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are 7 even numbers between 1 and 15, and if we count y as a conontant there are 21 consonants in the alphabet. There are therefore 7 * 21 * 3 = 441 possible 3-object collections containing an even number, a consonant, and one of the first three colors of rainbow. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the answer to this question we simply take the number of, even numbers multiplied by the number of letters in the alphabet multiplied by the number of colors which gives us the following equation. 7 * 26 * 7 = 1274 contain an even number, and to find the number of combinations that have a vowel in them we simply take 15 * 5 * 7 = 525 which is how many of these combinations contain a vowel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are 7 * 26 * 7 = 1274 collections which contain aneven number. There are 15 * 5 * 7 = 525 collections which contain avowel. 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 numberalso contain a vowel, and are therefore included in the525 collections containing vowels. If we add the 1274 and the 525 we are counting each of these even-number-and-vowelcollections twice. We can correct for this error by determining how manyof the collections in fact contain an even number AND avowel. 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 sum1274 + 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 theirintersection, and n(S) stands for the number of objects in the setS. As the rule is applied here, A is the set of collections containing aneven 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 needed to give more detail in my answer. A break down needed to be given and a I also needed to put in where I got that particular number. ------------------------------------------------ Self-critique Rating: ********************************************* 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: You must remember that in the leading question it states that there are 15 possibilities for the first ball chosen because in box one we have 15 balls. In the second box we have letter tiles with one letter of the alphabet per tile. In the third box we have colored rings one for each color of the rainbow with is ROY G BIV. Keeping this information in mind we have to consider there are 15 balls therefore 15 possibilities for the first ball chosen which then leaves 14 possibilities for the second ball chosen. We also know that there are a total of 26 tiles with letters and in the thired box there are a total of 7 rings. Knowing all of this information all we need to do at this point is simply multiply like so: 15 * 14 * 26 * 7 / 2 now with the number 2 the reason that it is in this equashion is because of the fact that this is a collection and the order of the numbers does not matter and therefore since the 2 represents the amount of choices made for the balls we simply use that number since it fell first. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: There are 15 possibilities for the first ball chosen, which leaves14 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 theobjects. 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 choicewould look different if ball 7 was placed before ball 12than 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 ischosen first. Since the two balls in any given collectioncould have been chosen in either of two orders, there are only half as many possible collections as there areordered choices. In this case, since the order in which the balls are chosen doesnt 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 dont understand what you mean by the /2 at the end of the first part if you say the order chosen doesnt matter. Why are you dividing by 2? Is that because you are picking two and then the order doesnt 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.