#$&* course Mth 152 004. Dice, trees, committees, number of subsets.
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Given Solution: There are two dice. Call one the 'first die' and the other the 'second die'. (Note that ‘die’ is the singular of ‘dice’). • It is possible for the first die to come up 3 and the second to come up 6. • It is possible for the first die to come up 4 and the second to come up 5. • It is possible for the first die to come up 5 and the second to come up 4. • It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* Question: `q002. In how many ways can we choose a committee of three people from a set of five people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can do this as a permutation. P (5, 3) 5! / [3! (5 - 3)! ] 5! / 3! * 2! 5 * 4 * 3 * 2 * 1 / (3 * 2 * 1) * (2 * 1) 5 * 4 / (2 * 1) 5 * 2 = 10 ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OKAY - need to remember it’s C not P ------------------------------------------------ Self-critique Rating: OKAY ********************************************* Question: `q003. In how many ways can we choose a president, a secretary and a treasurer from a group of 10 people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This can also be done as a permutation. It’s specific but not gender specific like the last question that was asked about president, secretaries and treasurers. We are looking for 3 positions to be filled out of 10 so called applicants and how many ways that can be done. P (10,3) 10!/ (10-3)! 10!/7! 10*9*8*7*6*5*4*3*2*1/7*6*5*4*3*2*1 10*9*8* = 720 ways to fill the three positions form 10 applicants. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* Question: `q004. In how many ways can we arrange six people in a line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want to arrange 6 people, but we don’t need to arrange them in any specific order. The formula is 6! = 6*5*4*3*2*1.There are 720 ways to arrange the people. Confidence rating: 3
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Given Solution: There are 6 ! = 720 possible orders in which to arrange six people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* Question: `q005. In how many ways can we rearrange the letters in the word 'formed'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the word ‘formed’ there are 6 letters. This question can be solved the same way the question before was. Rearranging the letters in no distinct order will still equal out to 720 different ways from 6! It doesn’t matter in this case that one is letters and one is dealing with people - it is only being ordered or unordered that matters. 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): OKAY ------------------------------------------------ Self-critique Rating: OKAY ********************************************* Question: `q006. In how many ways can we rearrange the letters in the word 'activities'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the word ‘activities’ there are 10 letters. These 10 letters do not have to be arranged in any specific order. We would solve by using 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800 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): Oh, I completely forgot about repeating letters! But to be honest I would of asked the same question as the student comment. Looking at the problem I just don’t assume to look for that or remember that because it seems sometimes there’s an exception where other times there is not. But looking at the solution 10 ! / ( 3 ! * 2 !), that does makes sense to me ------------------------------------------------ Self-critique Rating: 2
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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): OKAY - is my answer right??? ------------------------------------------------ Self-critique Rating: Okay