course Mth 152 Ûä¤R¢óü…Åx¢{½Ëܹ“‘®êx‹{ÓŒassignment #004
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18:19:21 `q001. Note that there are 9 questions in this assignment. In how many ways can we get a total of 9 when rolling two fair dice?
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RESPONSE --> 3 and 6 6 and 3 4 and 5 5 and 4 confidence assessment: 3
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18:19:56 There are two dice. Call one the 'first die' and the other the 'second die'. It is possible for the first die to come up 3 and the second to come up 6. It is possible for the first die to come up 4 and the second to come up 5. It is possible for the first die to come up 5 and the second to come up 4. It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3).
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RESPONSE --> Ok. self critique assessment: 3
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18:22:18 `q002. In how many ways can we choose a committee of three people from a set of five people?
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RESPONSE --> 1, 2, 3, 4, 5 1, 2, 3 1, 2, 4 1, 2, 5 1, 3, 4 1, 3, 5 1, 4, 5 2, 3, 4 2, 3, 5 2, 4, 5 3, 4, 5 confidence assessment: 3
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18:23:03 A committee when first chosen is understood to consist of equal individuals. The committee is therefore unordered, and we see that in choosing a committee of three people from a set of five people we are forming a combination of 3 people from among 5 candidates. The number of such combinations is C ( 5, 3) = 5 ! / [ 3 ! ( 5 - 3) ! ] = 5 ! / [ 3 ! * 2 ! ] = 5 * 4 * 3 * 2 * 1 / [ ( 3 * 2 * 1 ) * ( 2 * 1) ] = 5 * 4 / ( 2 * 1) = 5 * 2 = 10. There are 10 possible 3-member committees within a group of 5 individuals.
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RESPONSE --> Ok. self critique assessment: 3
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18:26:25 `q003. In how many ways can we choose a president, a secretary and a treasurer from a group of 10 people?
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RESPONSE --> P(10, 3) = 10! / 7! = 10 * 9 * 8 = 720 confidence assessment: 3
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18:27:12 This choice is ordered. The order of our choice determines who becomes president, who becomes secretary and who becomes treasurer. Since we are choosing three people from 10, and order matters, we are looking for the number of permutations of three objects chosen from 10. This number is P(10, 3) = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720.
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RESPONSE --> Ok. self critique assessment: 3
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18:27:48 `q004. In how many ways can we arrange six people in a line?
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RESPONSE --> 6! = 720 possibilities confidence assessment: 3
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18:28:01 There are 6 ! = 720 possible orders in which to arrange six people.
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RESPONSE --> Ok. self critique assessment: 3
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18:28:33 `q005. In how many ways can we rearrange the letters in the word 'formed'?
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RESPONSE --> 6! = 720 possibilities confidence assessment: 3
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18:29:01 There are six distinct letters in the word 'formed'. Thus we can rearrange the letters in 6 ! = 720 different ways.
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RESPONSE --> Ok. self critique assessment: 3
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18:32:47 `q006. In how many ways can we rearrange the letters in the word 'activities'?
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RESPONSE --> 10! / 6! confidence assessment: 3
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18:34:11 There are 10 letters in the word 'activities', but some of them are repeated. There are two t's and three i's. If we think of these 10 letters as being placed on letter tiles, there are 10! ways to rearrange the tiles. However, not all of these 10 ! ways lead to different words. For any spelling the three 'i' tiles can be arranged in 3 ! = 6 different ways, all of which spelled same word. And for any spelling the two 't' tiles can be arranged in 2 ! = 2 different ways. We must thus divide the 10! by 3! and by 2!, leading to the conclusion that there are 10 ! / ( 3 ! * 2 !) different spellings of the rearranged tiles.
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RESPONSE --> Ok. self critique assessment: 3
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18:35:34 `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph?
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RESPONSE --> P (10, 4) = 10 ! / 6 ! = 10 * 9 * 8 * 7 confidence assessment: 3
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18:36:27 We are arranging four people chosen from 10, in order. The number of possible arrangements is therefore P ( 10, 4) = 10! / ( 10-4)! = 10 ! / 6 ! = 10 * 9 * 8 * 7.
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RESPONSE --> Ok. self critique assessment: 3
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18:38:33 `q008. In how many ways can we get a total greater than 3 when rolling two fair dice?
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RESPONSE --> Only three ways to get less than 3 out of 36 overall possibilities 36 - 3 = 33 possibilities confidence assessment: 3
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18:39:28 It would not be difficult to determine the number of ways to get totals of 4, 5, 6, etc. However it is easier to see that there is only one way to get a total of 2, which is to get 1 on both dice; and that there are 2 ways to get a total of 3 (we can get 1 on the first die and 2 on the second, or vice versa). So there are 3 ways to get 3 or less. Since there are 6 possible outcomes for the first die and 6 possible outcomes for the second, there are 6 * 6 = 36 possible outcomes for the two dice. Of these 36 we just saw that 3 give a total of 3 or less, so there must be 36 - 3 = 33 ways to get more than 3.
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RESPONSE --> Ok. self critique assessment: 3
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18:42:17 `q009. A committee consists of 5 men and 7 women. In how many ways can a subcommittee of 4 be chosen if the number of men and women must be equal?
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RESPONSE --> C(5, 2) * C(7, 2) = 10 * 21 = 210 possibilities. confidence assessment: 3
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18:43:42 If the numbers of women and men are equal, then there must be 2 of each. Recall that a committee is regarded as unordered. If order doesn't matter there are C(5, 2) ways to choose 2 men out of 5, and C(7, 2) ways to choose 2 women out of 7. We have to choose 2 men AND we have to choose 2 women, so the Fundamental Counting Principal tells us that there are C(5, 2) * C(7, 2) = 10 * 21 = 210 possible subcommittees.
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RESPONSE --> Ok. self critique assessment: 3
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“Œ‹ÛšøÀzš·xÝ|MŽÒË”²W ©úzœIûÞ´ assignment #005 005. Binary probabilities Liberal Arts Mathematics II 06-08-2007
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18:48:43 `q001. Note that there are 10 questions in this assignment. List the possible outcomes if a fair coin is flipped 2 times.
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RESPONSE --> Heads - Heads Heads - Tails Tails - Heads Tails - Tails confidence assessment: 3
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18:49:27 There are 2 coins. Call one of them the first and the other the second coin. We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT. Thus there are 4 possible outcomes: HH, HT, TH and TT.
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RESPONSE --> Ok. self critique assessment: 3
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18:53:18 `q002. List the possible outcomes if a fair coin is flipped 3 times.
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RESPONSE --> Heads - Heads - Heads Heads - Heads - Tails Heads - Tails - Heads Heads - Tails - Tails Tails - Heads - Heads Tails - Heads - Tails Tails - Tails - Heads Tails - Tails - Tails confidence assessment: 3
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18:56:01 The possible results for the first 2 flips are HH, HT, TH and TT. We can obtain all possible results for 3 flips by appending either H or T to this list. We start out by writing the list twice: HH, HT, TH, TT HH, HT, TH, TT We then append H to each outcome in the first row, and T to each outcome in the second. We obtain HHH, HHT, HTH, HTT THH, THT, TTH, TTT Note that this process shows clearly why the number of possibilities doubles when the number of coins increases by one. With two coins we had 4 possible outcomes and with three coins we had 8 outcomes, twice as many as with two coins.
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RESPONSE --> Ok. self critique assessment: 3
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19:00:42 `q003. List the possible outcomes if a fair coin is flipped 4 times.
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RESPONSE --> HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT TTTT TTTH TTHT TTHH THTT THHT THHH confidence assessment: 3
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19:01:25 We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Then we append H to the front of one list and T to the front of the other: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins.
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RESPONSE --> Ok. self critique assessment: 3
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19:09:26 `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'?
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RESPONSE --> HHTT TTHH HTHT THTH HTTH THHT confidence assessment: 3
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19:10:47 The two 'heads' can occur in positions 1 and 2 (HHTT), 1 and 3 (HTHT), 1 and 4 (HTTH), 2 and 3 (THHT), 2 and 4 (THTH), or 3 and 4 (TTHH). These six possibilities can be expressed by the sets {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. Thus the possibilities are represented by sets of two numbers chosen from the set {1, 2, 3, 4}. When choosing 2 numbers from a set of four, there are 4 * 3 / 2 possible combinations. Since in this case it doesn't matter in which order the two positions are picked, this will be the number of possible outcomes with exactly two 'heads'. The number of possibilities is thus C(4, 2) = 6.
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RESPONSE --> Ok. self critique assessment: 3
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19:14:23 `q005. If a fair coin is flipped 7 times, how many of the outcomes contain exactly three 'heads'?
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RESPONSE --> C(7, 3) = 7 * 6 * 5 / 3! = 35 possibilities confidence assessment: 3
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19:16:35 The possible positions for the three 'heads' can be numbered 1 through 7. We have to choose three positions out of these seven possibilities, and the order in which our choices occur is not important. This is equivalent to choosing three numbers from the set {1, 2, 3, 4, 5, 6, 7} without regard for order. This can be done in C(7,3) = 7 * 6 * 5 / 3! = 35 ways. There are thus 35 ways to obtain 3 'heads' on 7 flips.
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RESPONSE --> Ok. self critique assessment: 3
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19:18:49 `q006. If we flip a fair coin 6 times, in how many ways can we get no 'heads'? In how many ways can we get exactly one 'head'? In how many ways can we get exactly two 'heads'?
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RESPONSE --> C(6, 0) = 1 C(6, 1) = 6 C(6, 2) = 15 confidence assessment: 3
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19:19:39 In how many ways can we get exactly three 'heads'?
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RESPONSE --> C(6, 3) = 20 confidence assessment: 3
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19:20:06 In how many ways can we get exactly four 'heads'? In how many ways can we get exactly five 'heads'?
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RESPONSE --> C(6, 4) = 15 C(6, 5) = 6 confidence assessment: 3
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19:20:30 In how many ways can we get exactly six 'heads'?
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RESPONSE --> C(6, 6) = 1 confidence assessment: 3
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19:21:30 In how many ways can we get exactly seven 'heads'?
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RESPONSE --> None confidence assessment: 3
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19:22:04 The number of ways to get no 'heads' is C(6,0) = 1. The number of ways to get exactly one 'head' is C(6,1) = 6. The number of ways to get exactly two 'heads' is C(6,2) = 15. The number of ways to get exactly three 'heads' is C(6,3) = 20. The number of ways to get exactly four 'heads' is C(6,4) = 15. The number of ways to get exactly five 'heads' is C(6,5) = 6. The number of ways to get exactly six 'heads' is C(6,6) = 1. These numbers form the n = 6 row of Pascal's Triangle: 1 6 15 20 15 6 1 See your text for a description of Pascal's Triangle. Note also that these numbers add up to 64, which is 2^6, the number of possible outcomes when a coin is flipped 6 times.
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RESPONSE --> Ok. self critique assessment: 3
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19:30:29 `q007. List all the subsets of the set {a, b}. Then do the same for the set {a,b,c}. Then do the same for the set {a,b,c,d}.
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RESPONSE --> { }, {a}, {b}, {a, b} { }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} { }, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a. d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d} confidence assessment: 3
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19:31:25 The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}. These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}. The subsets are therefore {}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}. We see that the number of subsets doubles when the number of elements in the set increases by one. This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows: To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips.
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RESPONSE --> Ok. self critique assessment: 3
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19:34:37 `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> 2^8 = 256 subsets confidence assessment: 3
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19:35:37 There are 2 possible subsets of the set {a}, the subsets being { } and {a, b}. The number doubles with each additional element. It follows that for a set of 2 elements there are 2 * 2 subsets (double the 2 subsets of a one-element set), double this or 2 * 2 * 2 subsets of a set with 3 elements, double this or 2 * 2 * 2 * 2 subsets of a set with 4 elements, etc.. There are thus 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256 subsets of the given set, which has 8 elements. This number is also written as 2^8. }{More generally there are 2^n subsets of any set with n elements.
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RESPONSE --> Ok. self critique assessment: 3
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19:36:42 `q009. How many 4-element subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> C(8, 4) = 8 *7 * 6 * 5 / (4 * 3 * 2 * 1) = 70 confidence assessment: 3
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19:38:24 To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice. The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70.
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RESPONSE --> Ok. self critique assessment: 3
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19:40:02 `q010. How many subsets of the set {a,b,c,d} contain 4 elements? How many subsets of the set {a,b,c,d} contain 3 elements? How many subsets of the set {a,b,c,d} contain 2 elements? How many subsets of the set {a,b,c,d} contain 1 elements? How many subsets of the set {a,b,c,d} contain no elements?
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RESPONSE --> C(4,4) = 1. C(4,3) = 4. C(4,2) = 6. C(4,1) = 4. C(4,0) = 1. confidence assessment: 3
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19:40:40 The number of 4-element subsets is C(4,4) = 1. The number of 3-element subsets is C(4,3) = 4. The number of 2-element subsets is C(4,2) = 6. The number of 1-element subsets is C(4,1) = 4. The number of 0-element subsets is C(4,0) = 1. We note that these numbers form the n = 4 row 1 4 6 4 1 of Pascal's Triangle, and that they add up to 2^4 = 16, the number of possible subsets of a 4-element set.
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RESPONSE --> Ok. self critique assessment: 3
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yô”ŸzT«²¶Œ¯æ³Èäõé̸ә« assignment #004 004. `query 4 Liberal Arts Mathematics II 06-08-2007
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19:59:41 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9, 6) = 9! / (6! (6-3)!) = 9*8*7 / (3*2*1) = 3*4*7 = 84
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20:00:08 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> Ok.
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20:02:36 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> C(9, 4) = 126 C(8, 4) = 70
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20:03:15 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> Ok.
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20:04:17 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> Fibonacci Numbers.
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20:04:54 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> Ok.
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20:07:18 Query 11.4.42 (x+y)^8
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RESPONSE -->
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20:07:50 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE -->
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20:08:33 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None
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20:08:56 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> Ok.
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ק¯£½ÿµÂøW™Vw¥ÊëÒ¿KÖ®—Â¥¨¾‚Ï„ assignment #005 005. `query 5 Liberal Arts Mathematics II 06-08-2007
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20:15:58 Query 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
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RESPONSE --> 30 ways to get different numbers
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20:16:53 ** On two fair dice you have 6 possible outcomes on the first and 6 on the second. By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes. We can list these outcomes in the form of ordered pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Of these 36 outcomes there are six that have the same number on both dice. It follows that the remaining 3 - 6 = 30 have different numbers. So there are 30 ways to get different numbers on the two dice. Note that your chance of getting different numbers is therefore 30 / 36 = 5/6 = .8333... .**
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RESPONSE --> Ok.
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20:21:30 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> C(52, 13) - 4
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20:22:11 ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13). There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit). Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands. Of the number of hands having more than one suit is C(52, 13) - 4. **
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RESPONSE --> Ok.
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20:27:39 11.5.20 # subsets of 12-elt set with from 3 to 9 elts? How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
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RESPONSE --> 3: C(12, 3) 4: C(12, 4) 5: C(12, 5) 6: C(12, 6) 7: C(12, 7) 8: C(12, 8) 9: C(12, 9) 2^12 = 4096 - 79*2 = 3938 sets between 3 and 9
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20:28:32 ** You need the number of subsets with 3 elements, with 4 elements, etc.. You will then add these numbers to get the total number of 3-, 4-, 5-, ., 9-element subsets. Start with a 3-e.ement subset. In a 12-element set, how many subsets have exactly three elements? You answer this by asking how many possibilities there are for the first element, then how many for the second, then how many for the third. You can choose the first element from the entire set of 12, so you have 12 choices. You have 11 elements from which to choose the second, so there are 11 choices. You then have 10 elements left from which to choose the third. So there are 12 * 11 * 10 ways to choose the elements. However, the order of a set doesn't matter. 3 elements could be ordered in 3! different ways, so there are 12 * 11 * 10 / 3! ways to choose different 3-element sets. This is equal to C(12,3). So there are C(12, 3) 3-elements subsets of a set of 12 elements. Reasoning similarly we find that there are C(12,4) ways to choose a 4-element subset. C(12,5) ways to choose a 5-element subset. C(12,6) ways to choose a 6-element subset. C(12,7) ways to choose a 7-element subset. C(12,8) ways to choose a 8-element subset. C(12,9) ways to choose a 9-element subset. We see that there are C(12,3) + C(12,4) + C(12,5) + C(12,6) + C(12,7) + C(12,8) + C(12,9) possible subsets with 3, 4, 5, 6, 7, 8 or 9 elements. Alternatively you can figure out how many sets have fewer than 3 or more than 9 elements. There are C(12, 0) + C(12, 1) + C(12, 2) = 1 + 12 + 66 = 79 sets with fewer than 3 elements, and C(12, 10) + C(12, 11) + C(12, 12) = 66 + 12 + 1 = 79 sets with more than 3. Since there are 2^12 = 4096 possible subsets of a 12-element set there are 4096 - 79 - 79 = 3938 sets with between 3 and 9 elements. **
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RESPONSE --> Ok.
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20:33:10 11.5.30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
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RESPONSE --> 40*4*4*4*4
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20:35:06 ** There are 9 choices for the denomination of the lowest card in a straight, which gives 36 cards that could be the low card. However but if aces can be high or low there are 40. There are then four choices for the next-higher card, four for the next after that, etc., giving 40*4*4*4*4 possibilities. **STUDENT COMMENT: I don 't understand this one . Idon't see where you get the 9 from. INSTRUCTOR RESPONSE: Cards run from 2 through 10, then the four face cards, then the ace. You need five consecutive cards to make a straight. The highest possible straight is therefore 10, Jack, Queen, King and Ace. The lowest is 2, 3, 4, 5, 6. The lowest card of the straight can be any number from 2 through 10. That is 9 possibilities. In some games the ace can be counted as the low card, below the 2, as well as the high card. In that case there would be one more possibility for a straight, which could then consists of denominations 1, 2, 3, 4, 5. *&*&
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RESPONSE --> Ok.
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20:37:20 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE --> 6 * 3 = 18 possibilities
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20:39:16 ** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities. A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25. A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650. Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement. The fundamental counting principle is the key here. STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations. You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
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RESPONSE --> Ok.
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20:40:39 Query 11.5.48 # 3-digit counting #'s without digits 2,5,7,8?
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RESPONSE --> 5*6*6 = 180 possibilities
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20:41:26 ** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
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RESPONSE --> Ok.
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20:41:52 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None
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