#$&* course MTH 152 Please note that I am still working on this query, namely the last problem. But I need to get back on track ASAP on the third question (11.3.43/42) If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution:: ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices, respectively. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 25 ! / ( (25 - 5) !) is P(25, 5). We use permutations because in this case, there are 5 different prizes so the order in which the students are chosen makes a difference in the final outcome. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Please explain your thought process/reasoning throughout the following passage. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 25 ! / ( (25 - 5) !) is P(25, 5). ********************************************* question: Is repetition allowed in this situation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Since there are five prizes to be given to each of the 5 students chosen from a collection of 25; no, repetition is not allowed. Unless of course you plan to take a student’s prize away from him/her which, depending on the age-level, could be considered ill-mannered and morally reprehensible. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok. Noted that one cannot either take a prize away or give the same prize to more than one student. ------------------------------------------------ Self-critique Rating: question 10.3.30 (previously 11.3.30) 3-letter monogram all letters different YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Where C(25, 2) = 25 * 24 / 2! = 300 possible combinations, I would paraphrase that - C(25,2) is the same as saying (25 possible options, 2 choices being made) because the first two letters CANNOT be ‘z’ therefore the total number of the options is reduced by 1. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** We are choosing 3 different letters, and since the monogram will be different if you change the order of the letters, we can say that order definitely applies. If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices. We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26 possible objects. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However in this question the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first). So there are only 25 * 24 = 600 possibilities. ** MODIFIED SOLUTION: I believe in the original solution that I overlooked the requirement that the letters be in alphabetical order. Z is the last letter, so as long as the other two are chosen from the first 25 letters of the alphabet, it will be possible to construct the monogram. Any combination of two of the 25 remaining letters can be used. Once the combination is selected, the letters will then be put into alphabetical order. There are C(25, 2) = 25 * 24 / 2! = 300 possible combinations, so there are 300 possible monograms with the letters distinct and fin order. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): See my comment above. For some apparent reason, I’m having trouble comprehending the material with this assignment so far. ------------------------------------------------ Self-critique Rating: ok question (previously 11.3.43 /& 42) divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - Note that the sum of 3 -> 7 is 25, our total. - I claim that order doesn’t matter here, as to which students are put into what group. - The number of students within a group is, for now, ignored and instead the number of groups is considered, being 5 different groups. - Or, another option, is to consider C (25, 3) C (25, 4) C (25, 5) C (25, 6) C (25, 7) - Note that since the claim that order does NOT matter here, conclude that - C (n, r) = n! / [(n-r)! *r!] - C(25, 3) = 25! / [ (25 - 3)! * 3 !] - C(25, 4) = 25! / [ (25 - 4)! * 4 !] - C(25, 5) = 25! / [ (25 - 5)! * 5 !] - C(25, 6) = 25! / [ (25 - 6)! * 6 !] - C(25, 7) = 25! / [ (25 - 7)! * 7 !] - ************ One has noted that, in temporary lapse, the remainder of students left depend on the number of group that they are placed in. For example, - - C (25, 3) C (25, 4) C (25, 5) C (25, 6) C (25, 7) - Becomes - C (25, 3) C (22, 4) C (18, 5) C (13, 6) C (7, 7) Therefore, using the FCP, we can state that 25* 22* 18 * 13 * 7 = 900,900 ways to group 25 students into five separate groups of 3,4,5,6, and 7 students. ****************************************************************************** Scrap everything above the asterisks. For now, I have deduced the following: - Since order doesn’t matter, I first follow the path the question leads, to divide the students up into five groups. - C (25, 3) = 25! / [ (25 - 3)! * 3! = 2500 C (22, 4) = 22! / [ (22 - 4)! * 4! = 7315 C (18, 5) = 18! / [ (18 - 5)! * 5! = 8568 C (13, 6) = 22! / [ (13 - 6)! * 6! = 1716 C (7, 7) = 7! / [ (7 - 7)! * 7! = 1 - Taking the sum from the above solutions = 19,900 combinations. As you had mentioned in the Given Solution, since order doesn’t matter, one can reverse the numbers. So next I put this to the test. - C (25, 7) = 25! / [ (25 - 7)! * 7! = 480700 - C (18, 6) = 18! / [ (18 - 6)! * 6! = 18564 - C (12, 5) = 12! / [(12 - 5)! * 5! = 792 - C (7, 4) = 7! / [ ( 7-4) ! * 4! ] = 35 - C (3, 3) = 3! / [ (3 -3)! * 3! ] = 1 - Again taking the sum from the above solutions = 500,092 So something is awry here. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** We can choose the groups in any order we wish. Each group chosen is chosen without regard for order. If we choose to begin by making the group of 3, there are 25 students available when we begin to select our group, so there are C(25, 3) possible choices. If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices. If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices. If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices. If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices. The Fundamental Counting Principle says you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group. So get have C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) ways to complete the grouping. Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do complete the groupings. The question is, would this make a difference in the final result? To find out we compare the two results C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3). If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified, both turn out to have exactly the same numbers in their numerators, and the same numbers in their denominators. As a result, they both end up in the same form 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I obviously derailed somewhere with this. Please provide step-by-step guidance so that I may get back on track.
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Given Solution: ** C(n, r) is the number of ways of choosing r objects out of n available objects, without regard for order. C(n, 0) is therefore the number of ways to choose 0 objects from among n objects. No matter what n is, there is exactly one way to do this, which is to choose nothing. Thus C(n, 0) is always equal to zero. As another example: There are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) is the number of ways to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us the same result: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!