#$&* course mth 152 004. Dice, trees, committees, number of subsets.
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Given Solution: There are two dice. Call one the 'first die' and the other the 'second die'. (Note that ‘die’ is the singular of ‘dice’). • It is possible for the first die to come up 3 and the second to come up 6. • It is possible for the first die to come up 4 and the second to come up 5. • It is possible for the first die to come up 5 and the second to come up 4. • It is possible for the first die to come up 6 and the second to come up 3. These are the only possible ways to get a total of 9. Thus there are 4 ways. We can represent these 4 ways as ordered pairs: (3,6), (4, 5), (5, 4), (6, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. In how many ways can we choose a committee of three people from a set of five people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C(5,3) = 5 ! / [ 3 ! (5-3)!] = 5! / [ 3! * 2] 5 * 4 * 3 * 2 * 1 / [ (3 * 2 * 1) * (2 * 1) ] 5 * 4 / (2*1) = 5*2 =10 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: P(10,3) = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This choice is ordered. The order of our choice determines who becomes president, who becomes secretary and who becomes treasurer. Since we are choosing three people from 10, and order matters, we are looking for the number of permutations of three objects chosen from 10. This number is • P(10, 3) = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720. STUDENT COMMENT: I am so lost on the formulas, I have no clue which one to use where. These last two questions looked just a like to me. INSTRUCTOR RESPONSE: Between these two problems, order matters on one, it doesn't matter on the other. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. In how many ways can we arrange six people in a line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6! = 720 ways to arrange six people in a line. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 6 ! = 720 possible orders in which to arrange six people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. In how many ways can we rearrange the letters in the word 'formed'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are six different letters in the word. We can arrange them in 6 ! = 720 different ways. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are six distinct letters in the word 'formed'. Thus we can rearrange the letters in 6 ! = 720 different ways. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. In how many ways can we rearrange the letters in the word 'activities'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For spelling the three I tiles can be arranged in 3 ! =6 different ways For spelling any two t tiles can be arranged in 2 ! = 2 different ways 10 ! / ( 3! * 2 !) different words are possible. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: P ( 10,4 ) = 10 ! / ( 10-4 ) ! = 10 ! / 6 ! = 10 * 9 * 8 * 7 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We are arranging four people chosen from 10, in order. The number of possible arrangements is therefore • P ( 10, 4) = 10! / ( 10-4)! = 10 ! / 6 ! = 10 * 9 * 8 * 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. In how many ways can we get a total greater than 3 when rolling two fair dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Three ways of getting a three when rolling two dice. There are 6 possible outcomes for the first die and 6 possible outcomes for the second die so 6* 6 = 36 possible outcomes for the two dice 36 - 3 = 33 ways to get more than 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: e `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: There must be two men and two women for it to be equal. C(5,2) * c(7,2) = 10 * 21 = 210 possible subcommittees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the numbers of women and men are equal, then there must be 2 of each. Recall that a committee is regarded as unordered. If order doesn't matter there are C(5, 2) ways to choose 2 men out of 5, and C(7, 2) ways to choose 2 women out of 7. We have to choose 2 men AND we have to choose 2 women, so by the Fundamental Counting Principal there are • C(5, 2) * C(7, 2) = 10 * 21 = 210 possible subcommittees. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: e `q009. A committee consists of 5 men and 7 women. In how many ways can a subcommittee of 4 be chosen if the number of men and women must be equal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There must be two men and two women for it to be equal. C(5,2) * c(7,2) = 10 * 21 = 210 possible subcommittees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: e `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: There must be two men and two women for it to be equal. C(5,2) * c(7,2) = 10 * 21 = 210 possible subcommittees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!