course Mth 152 I}vϓWassignment #001
......!!!!!!!!...................................
19:04:50 `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.
......!!!!!!!!...................................
RESPONSE --> abc acb bac bca cab cba confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:07:20 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.
......!!!!!!!!...................................
RESPONSE --> I understand why I got this right. It helped that I used alphabetical order self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:10:19 `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'.
......!!!!!!!!...................................
RESPONSE --> aaa aab aac aba aca bbb bba bbc bab bcb ccc cca ccb cac cbc abc acb bac bca cab cbc confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:12:27 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'.
......!!!!!!!!...................................
RESPONSE --> I missed 6 of them here. I left out aab aac bba bbc cca ccb self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:16:34 `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'?
......!!!!!!!!...................................
RESPONSE --> abc bca cab 3 for the 1st and 2nd therfore we have 3 choices for the 2 letter word formed by the first 3 letters 3 choices for the 3rd confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:00:43 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.
......!!!!!!!!...................................
RESPONSE --> I misunderstood. I do understand that multiplying the the chioces for the 1st, 2nd, and 3rd letters will give you the number possible. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:01:21 `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?
......!!!!!!!!...................................
RESPONSE --> a, b, and c yes confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:03:31 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.
......!!!!!!!!...................................
RESPONSE --> I understand that the that the first 2 letters could be ab, ac, ba, bc, ca, or cb and that the 3rd letter is determent by the 1st 2 self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:07:12 `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'?
......!!!!!!!!...................................
RESPONSE --> 3 for the 1st letter 3 fot the 2nd 9 to form a 2 letter word there are 3 choices left for the third letter confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:08:28 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.
......!!!!!!!!...................................
RESPONSE --> I did not multiply it out, but I understand how to, and I did answer that there is a possibility of 3 letters for each letter in the word. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:12:28 `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'?
......!!!!!!!!...................................
RESPONSE --> 1st letter= 4 2nd letter =3 There would be 12 choices for a two letter word 4*3= 12 3rd letter= 2 4*3*2=24 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:12:56 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'.
......!!!!!!!!...................................
RESPONSE --> I understand this concept as shown in my answer. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:21:36 `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?
......!!!!!!!!...................................
RESPONSE --> 4*3*2*1= 24 abcd abdc acbd acdb adac adca bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc dacb dbac dbca dcab dcba yes confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:22:02 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.
......!!!!!!!!...................................
RESPONSE --> I understand why I got this correct self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:26:29 `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?
......!!!!!!!!...................................
RESPONSE --> 15* 26* 7= 2730 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:27:29 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.
......!!!!!!!!...................................
RESPONSE --> I understand the multiplying the numbers of objects in each box gives you the solution. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:28:37 `q009. For the three boxes of the preceding problem, how many of the possible 3-object collections contain an odd number?
......!!!!!!!!...................................
RESPONSE --> 2730/ 2= 1365 since every other number is odd. confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:34:08 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.
......!!!!!!!!...................................
RESPONSE --> I understand that the only box containing numbered objects was the first which 8 of those are odd. Any of the choices in the other two have no impact therefore you would multiply: 8* 26* 7= 1456 self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:36:04 `q010. For the three boxes of the preceding problem, how many of the possible collections contain an odd number and a vowel?
......!!!!!!!!...................................
RESPONSE --> there are 8 odd numbers in the 1st box and 5 vowels (if you don't include y) therfore: 8*5*7= 280 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:36:40 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.
......!!!!!!!!...................................
RESPONSE --> I understand this concept. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:37:59 `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?
......!!!!!!!!...................................
RESPONSE --> 7*21*3= 441 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:38:17 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.
......!!!!!!!!...................................
RESPONSE --> I understand why I got the correct solution self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:40:12 `q012. For the three boxes of the preceding problem, how many of the possible collections contain an even number or a vowel?
......!!!!!!!!...................................
RESPONSE --> 12* 7= 84 because 5 vowels plus 7 even numbers is 12 confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:42:17 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.
......!!!!!!!!...................................
RESPONSE --> Not sue what happend, but I can't see anything but the send file on this one. self critique assessment:
.................................................
......!!!!!!!!...................................
20:45:31 `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?
......!!!!!!!!...................................
RESPONSE --> 15* 2730=40950 because you take 15*14*13 to get the number of possible (2730) outcomes of the 3 balls confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:51:08 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.
......!!!!!!!!...................................
RESPONSE --> When I gave my solution I don't know what I was thinking when I multiplied the 2730 by the 15, but I didn't mean to do that. However, I am a little confused on why there are fewer outcomes for those tossed into a container. Why do you divide 15* 14* 13 by six. self critique assessment: 1
.................................................