course Mth 152 ?X?{q??????~?????assignment #004
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00:42:01 `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 --> 2 confidence assessment: 1
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00:43:11 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 --> I was not sure if I was to count rolls according to the specific die. self critique assessment: 2
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00:51:29 `q002. In how many ways can we choose a committee of three people from a set of five people?
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RESPONSE --> 5!/(3!(5-3)!) = 10 confidence assessment: 2
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00:52:06 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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00:53:37 `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 --> 10!/(10-3)! = 720 confidence assessment: 2
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00:53:43 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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00:54:52 `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 confidence assessment: 2
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00:54:58 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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00:56:08 `q005. In how many ways can we rearrange the letters in the word 'formed'?
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RESPONSE --> 6! - 1 = 719 confidence assessment: 2
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00:56:24 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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01:02:12 `q006. In how many ways can we rearrange the letters in the word 'activities'?
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RESPONSE --> 10!/(3!*2!) = 302400 confidence assessment: 2
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01:02:24 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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01:03:50 `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 --> 10!/(10-4)! = 5040 confidence assessment: 2
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01:04:09 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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01:07:59 `q008. In how many ways can we get a total greater than 3 when rolling two fair dice?
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RESPONSE --> 3 confidence assessment: 1
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01:08:57 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 put down how many rolls would give less than or equal to 3. self critique assessment: 2
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01:17:14 `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 --> (5c2)*(7c2) = 210 confidence assessment: 2
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01:17:25 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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