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 #$&*: ********************************************* Question: `q002. In how many ways can we choose a committee of three people from a set of five people? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5!/( 3! (3-5)! )=5!/3!*2!= 5*4*3*2*1/(3*2*1)*(2*1)= 5*4/2*1= 20/2=10 10 possibilities.
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Given Solution: A committee when first chosen is understood to consist of equal individuals. The committee is therefore unordered. In choosing a committee of three people from a set of five people we are forming a combination of 3 people from among 5 candidates. The number of such combinations is C ( 5, 3) = 5 ! / [ 3 ! ( 5 - 3) ! ] = 5 ! / [ 3 ! * 2 ! ] = 5 * 4 * 3 * 2 * 1 / [ ( 3 * 2 * 1 ) * ( 2 * 1) ] = 5 * 4 / ( 2 * 1) = 5 * 2 = 10. There are 10 possible 3-member committees within a group of 5 individuals. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* 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: 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q004. In how many ways can we arrange six people in a line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6*5*4*3*2*1=720
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Given Solution: There are 6 ! = 720 possible orders in which to arrange six people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q005. In how many ways can we rearrange the letters in the word 'formed'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 letters, 6*5*4*3*2*1=720
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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 #$&*: ********************************************* Question: `q006. In how many ways can we rearrange the letters in the word 'activities'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 letters, 2 t’s, 3 I’s, =10!=3,628,800 10!/(3!*2!) = 10!/3*2*1)*(2*1) 10*9*8*7*6*5*4*3*2*1/3=1,209,600 possibilities.
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Given Solution: There are 10 letters in the word 'activities', but some of them are repeated. There are two t's and three i's. If we think of these 10 letters as being placed on letter tiles, there are 10! ways to rearrange the tiles. However, not all of these 10 ! ways spell different words. For any spelling the three 'i' tiles can be arranged in 3 ! = 6 different ways, all of which spell same word. And for any spelling the two 't' tiles can be arranged in 2 ! = 2 different ways. We must thus divide the 10! ways to arrange the ten tiles by the 3! ways in which the three i tiles might be ordered, and then divide this result by the 2! ways in which the three t tiles might be ordered, leading to the conclusion that 10 ! / ( 3 ! * 2 !) different ‘words’ are possible.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q007. In how many ways can we line up four people, chosen from a group of 10, for a photograph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10! / ( 10-4)! = 10!/6! = 10*9*8*7=5,040 ways.
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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 #$&*: ********************************************* Question: `q008. In how many ways can we get a total greater than 3 when rolling two fair dice? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6*6=36 36-3=33 ways to get a total greater than 3.
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Given Solution: It would not be difficult to determine the number of ways to get totals of 4, 5, 6, etc. However it is easier to see that there is only one way to get a total of 2, which is to get 1 on both dice; and that there are 2 ways to get a total of 3 (we can get 1 on the first die and 2 on the second, or vice versa). So there are 3 ways to get a total of 3 or less when rolling two dice. Since there are 6 possible outcomes for the first die and 6 possible outcomes for the second, there are 6 * 6 = 36 possible outcomes for the two dice. Of the 36 total possible outcomes, we just saw that 3 give a total of 3 or less, so there must be 36 - 3 = 33 ways to get more than 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* 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:
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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. "