#$&* course MTH 152 1:30 PM on 6/20/13 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: There are two dice. Call one the 'first die' and the other the 'second die'. (Note that ‘die’ is the singular of ‘dice’). It is possible for the first die to come up 3 and the second to come up 6. It is possible for the first die to come up 4 and the second to come up 5. It is possible for the first die to come up 5 and the second to come up 4. It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Confidence Rating is 2 to reflect the lack of finding the answer by mathematical means. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. In how many ways can we choose a committee of three people from a set of five people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - C(collection of 5 people, 3 chosen) C(5,3) = 5 ! / [ (5-3) * 3!] = 5 *4 * 3 *2 * 1 / (2 * 1) * 3 *2 *1 = 5 *4 / 2 *1 = 5/1 * 4/2 = 5 * 2 = 10. Therefore there are 10 possible combinations when choosing three people from a set of five. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A committee when first chosen is understood to consist of equal individuals. The committee is therefore unordered. 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. STUDENT COMMENT: I really need to get this formula down!!! When are we supposed to use each formula?? INSTRUCTOR RESPONSE Combinations when order doesn't matter; permutations when it does. There's more, and your book states all the conditions. There are 10 possible 3-member committees within a group of 5 individuals. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I find myself in the odd circumstance with being more comfortable dealing with billiard balls, lettered tiles, and colored rings than I am people. Aside from the humor, everything seems good. ------------------------------------------------ Self-critique Rating: ok
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Given Solution: 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. STUDENT COMMENT: I am so lost on the formulas, I have no clue which one to use where. These last two questions looked just a like to me. INSTRUCTOR RESPONSE: Between these two problems, order matters on one, it doesn't matter on the other. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok. I have a little more confidence in myself since I know these formulas by heart and when to use them; and these students apparently did not. I hope, though, they eventually got it down. ------------------------------------------------ Self-critique Rating: Ok. See comment above. ********************************************* Question: `q004. In how many ways can we arrange six people in a line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 6 *5 *4 *3 *2 *1 = 30 * 12 * 2 = 30 * 24 = 720 ways one can arrange six people in a line. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 6 ! = 720 possible orders in which to arrange six people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Good to go. ------------------------------------------------ Self-critique Rating: Ok. ********************************************* Question: `q005. In how many ways can we rearrange the letters in the word 'formed'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - ‘formed’ contains F-O-R-M-E-D = six letters, so this is 6! Which is again 720. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are six distinct letters in the word 'formed'. Thus we can rearrange the letters in 6 ! = 720 different ways. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. In how many ways can we rearrange the letters in the word 'activities'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Activities has A-C-T-I-V-I-T-I-E-S = 10 letters, so 10! - - 10 * 9 * 8 * 7 * 6 *5 *4 * 3* 2* 1 = 90 * 56 * 30 * 12 * 2 = 90 * 56 * 720 = 5040 * 720 = 3,648,800 - I should note that there are repetitions within this word, so I must change my previous answer to accommodate. - There are 10 letters, but the letters ‘t’ and ‘i’ will repeat themselves twice. - {A, C, E, I, S, T, V} will be the main set where I = is repeated 3 times - T = is repeated 2 times - confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 spell different words. For any spelling the three 'i' tiles can be arranged in 3 ! = 6 different ways, all of which spell same word. And for any spelling the two 't' tiles can be arranged in 2 ! = 2 different ways. We must thus divide the 10! ways to arrange the ten tiles by the 3! ways in which the three i tiles might be ordered, and then divide this result by the 2! ways in which the three t tiles might be ordered, leading to the conclusion that 10 ! / ( 3 ! * 2 !) different ‘words’ are possible.. STUDENT COMMENT: I didn’t think about this because no where did it say the tiles cannot be repeated. And it really doesn’t say the new arrangement needs to make a new word INSTRUCTOR RESPONSE: The example invokes 10 tiles. Any rearrangement of the 10 tiles forms a word, though most don't form a word in any known language. If the 10 tiles all contain different letters, then every different rearrangment of the tiles forms a different word, and the number of possible words is equal to the number of possible reorderings of the tiles. If two or more of the tiles contain the same letter, then a different ordering of the tiles doesn't necessarily form a different word. In this example there are three i's, on three different tiles. If you switch two of those tiles, then you have a different arrangement of the actual tiles but the word remains the same. So there are more arrangements of tiles than there are different words. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 210 different ways. Since the order doesn’t matter, confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. In how many ways can we get a total greater than 3 when rolling two fair dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - 33 possibilities. At first I had listed them out, by hand, since it wouldn’t be too time consuming as well as reinforce comprehension of your Given Solution. - (1,3) (2,2) (3, 1) (4, 1) (5, 1) (6, 1) - (1, 4) (2, 3) (3, 2) (4, 2) (5, 2) (6, 2) - (1, 5) (2, 4) (3, 3) (4, 3) (5, 3) (6, 3) - (1,6) (2, 5) (3, 4) (4, 4) (5, 4) (6, 4) - (2, 6) (3, 5) (4, 5) (5, 5) (6, 5) - (3, 6) (4, 6) (5, 6) (6, 6) Note that the possibilities constrained only the rolls that would be greater than 3, so (1,1) , (1, 2, and (2, 1) do not count. So thus 36 - 3 = 33 possible rolls greater than 3. I also note for future reference (and to save time) that one would simply use the FCP for the number of possible options (sides on the die, 6) 6 * 6 = 36 totals possible, minus the three options listed above that are less than 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 a total of 3 or less when rolling two dice. 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 the 36 total possible outcomes, 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. STUDENT QUESTION First you said there are two ways to get three (we can get 1 on the first die and 2 on the second, or vice versa). Then you say in the bullet there is three ways. Which one is it? INSTRUCTOR RESPONSE There are two ways to get a total of 3. There are three ways to get a total of 3 or less. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Noted the reminder of how handy the Fundamental Counting Principle is in this case. Saves time, and brain power. Yet I was compelled to check myself anyway. ------------------------------------------------ Self-critique Rating: Ok. ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - So we have 5 men, and 7 women which gives us a total of 12 people in the entire committee. - My first reaction would be to list 6 men and 6 women, but we only have 5 men in the example. So unless we pull a stranger into the committee, this won’t work (the parameters don’t list abducting strangers of either gender, so that’s out) - I check below to see your methods. - I note that 2 members of each gender must be used. - So this means C(5, 2) * C(7, 2): - C(5, 2) = 5! / [ ( 5- 2) ! * 2 !] = 5! / 3 ! * 2 ! = 5 * 4 / 1* 2 = 5/1 * 4/2 = 5 * 2 = 10 - C(7,2) = 7! / [ (7 - 2)! * 2!] = 7! / 5! * 2! = 7 * 6 / 1 * 2 = 7/1 * 6/2 = 7 * 3 = 21 - 10 * 21 = 210 ways to form a sub-committee. *** Why must one choose 2? Why not 3, 4, 5 from each group? ***
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Given Solution: 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 by the Fundamental Counting Principal there are C(5, 2) * C(7, 2) = 10 * 21 = 210 possible subcommittees. ********************************************* Question: `q010. In how many ways can we get a total of 10 or more when rolling two dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Without listing the possibilities, I try to solve this using only mathematics. - I note that there are a total of 6 * 6 = 36 possibilities total. - To get a total of 10 or more, one must roll at least a (4, 6) or (6, 4), the order matters not. - So there are at least 2 ways to roll a 10. Since the parameters list 10 or more, this counts as one of the ways to roll 10 or greater. - There are 30 ways to roll less than 10. - Thus, the number of ways to roll less than 10 subtracted from the total number of rolls - 36 - 30 = 6 ways to roll a number greater than or equal to 10. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q011. In choosing 5 balls, without replacement, from 15 numbered balls, in how many ways can we obtain a collection of balls that includes three single-digit numbers and two double-digit numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Consider the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} - Of the set, only {1, 2, 3, 4, 5, 6, 7, 8, 9} = 9 balls have single-digit numbers. - Of the set, only {10, 11, 12, 13, 14, 15} = 6 balls have double-digit numbers. - I would theorize that I need to work out the product of C(9, 3) and C(6, 2) - C(9,3) = 9! / [ (9 - 3)! * 3! = 9! / 6! * 3! = 9 * 8 * 7 / 3 * 2 * 1 = 3 * 4 * 7 = 84. - C(6,2) = 6! / [ (6 - 2)! * 2! = 6! / 4! * 2! = 6 * 5 / 2 * 1 = 3 * 5 = 15 - Thus C(9, 3) * C(6,2) = 1260 ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: Provide any direction/assistance as necessary. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!