course Mth 152 DԽxxZޕassignment #004 004. Dice, trees, committees, number of subsets. Liberal Arts Mathematics II 02-23-2008
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09:18:54 `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 --> (6,3), (5,4), (3,6), (4,5) Four ways confidence assessment: 2
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09:30:04 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 --> That is how I worked it self critique assessment: 2
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09:35:19 `q002. In how many ways can we choose a committee of three people from a set of five people?
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RESPONSE --> Since it doesn't say that order matters, you would use combination. 5! / (3! (5-3)!) = 5! / (3! * 2!) = 5*4*3*2*1 / ((3*2*1) 2*1) = 5*4 / 2*1 = 20/2 = 10 10 ways confidence assessment: 2
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09:36:22 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 --> That is the answer that I came up and that is how I worked it. I thought that you would use the combination since it didn't specify that order mattered. self critique assessment: 2
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09:40:04 `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 --> Since the order does matter you would use permutations. 10! / (10 - 3)! = 10! / 7! = 10*9*8 = 720. 720 ways to choose confidence assessment: 2
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09:43:37 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 --> I was hoping that order mattered since you have to choose a President, then a secretary and then the treasurer self critique assessment: 2
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10:16:09 `q004. In how many ways can we arrange six people in a line?
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RESPONSE --> 6*5*4*3*2*1 = 720 ways to arrange them confidence assessment: 2
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10:16:47 There are 6 ! = 720 possible orders in which to arrange six people.
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RESPONSE --> That is how I worked it. I actually remembered how to work it. self critique assessment: 2
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10:22:13 `q005. In how many ways can we rearrange the letters in the word 'formed'?
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RESPONSE --> 6 letters - 6*5*4*3*2*1 = 720 confidence assessment: 2
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10:22:32 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 --> That is how I worked it too. self critique assessment: 2
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10:23:16 `q006. In how many ways can we rearrange the letters in the word 'activities'?
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RESPONSE --> 10*9*8*7*6*5*4*3*2*1 = 3628800 ways confidence assessment: 2
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10:25:45 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 --> I didn't think about the letters being repeated but I see now what I did wrong and to look at them as letters on tiles. self critique assessment: 2
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10:37:49 `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 --> Since it doesn't say that order is important and can not be repeated: 10! / 4! (10-4)! = 10! / (4! * 6!) = 10*9*8*7*6*5*4*3*2*1 / (4*3*2*1 * 6*5*4*3*2*1 ) = 10*9*8*7 / 4*3*2*1 = 5040/24 = 210 confidence assessment: 2
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10:38:35 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 --> I didn't do it as in order, I didn't realzie that order mattered so I used the combination method. self critique assessment: 2
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10:39:49 `q008. In how many ways can we get a total greater than 3 when rolling two fair dice?
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RESPONSE --> could get 6 possiblities for the first dice and 6 possiblities for the second dice. 6 * 6 = 36 ways confidence assessment: 2
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10:40:23 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 --> I forgot to subtract the 3 from the answer since it has to be greater than 3. self critique assessment: 2
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10:45:00 `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 --> The subcommittee has to be 2 men and 2 women 2 out of 5 men 2 out of 7 women 7!/(7-2)! = 7! / 5! = 7*6 = 42 5!/(5-2)! = 5! / 3! = 5*4 = 20 42 + 20 = 62 ways confidence assessment: 2
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10:47:43 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 --> I didn't even come close. I used the permutation method instead of the combination method. I realize now that the combination method would have been used because the order doesn't matter. self critique assessment: 2
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