course Mth 152
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Given Solution:: ** Starting with 10! / [ 4! * (10-4) ! ], we replace (10 – 4) by 6 to get 10! / ( 4! * 6! ). Writing out the factorials our expression becomes 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] . The numerator and denominator could be multiplied separatedly then divided out but it's easier and more instructive to divide out like terms. Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by the same expression ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves us 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1). Every factor of the denominator divides into a number in the numerator without remainder: • Divide 4 in the denominator into 8 in the numerator, • divide 3 in the denominator into 9 in the numerator and • divide 2 in the denominator into 10 in the numerator and you end up with • 5 * 3 * 2 * 7 = 210. NOTE ON WHAT NOT TO DO: You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divide 3628800 by 16480. But that process would lose accuracy and be ridiculously long for an expression like 100 ! / ( 30! * 70!). For this calculation the numbers involved would be hundreds of digits long. Much better to simply divide out like factors until the denominator goes away, as it always will with expressions of this form. (form n ! / (r ! * (n – r)! ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, so I made it to: 10*9*8*7/4*3*2*1 But I didn’t realize that I could divide the denominators into the numerators. It makes more sense to do that, it just didn’t cross my mind. question 11.2.31 (10th edition #25) 3 switches in a row; fund count prin to find # of possible settings ********************************************* Your Solution: 3 choices for the first switch * 2 for the 2nd * 1 for the third = 6 total possible settings ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 2
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Given Solution:: ** There are two possible settings for the first switch, two for the second, two for the third. The setting of one switch is independent of the setting of any other switch so the fundamental counting principle holds. There are therefore 2 * 2 * 2 = 8 possible setting for the three switches. COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways. INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. No single choice has 3 possibilities. Also the setting of each switch is independent of the settings of the others, so the number of possibilities on one choice is independent of the number of possibilities on the other. Each of the 3 choices therefore has 2 possibilities. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t realize that they were on/off switches. I thought that it was the placement of three individual switches. Once I looked at the problem in the textbook, I understood the question and how the given answer was found. (Because each of the three switches could only be on or off, there were only two choices for each) question 11.2.31 (10th edition #27) If no two adjacent switches are off why does the fundamental counting principle not apply? ********************************************* Your Solution: Because we cannot say that: 2 for the first *1 for the second *2 for the third because it could be the 1st or 3rd light switch that was off. It doesn’t specify and therefore we do not know where to place the different choices. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
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Given Solution:: ** The reason the principle doesn’t apply is that the Fund. Counting Principle requires that the events be independent. Here the state of one switch influences the state of its neighbors (neither neighbor can be the same as that switch). So the choices are not independent. The Fund. Counting Principle requires that the events be independent. Since this is not the case the principle does not apply. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand why the principle doesn’t apply, but my reasoning doesn’t sound as good as the given solution. question 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}? ********************************************* Your Solution: 3*3*2= 18 (the first and second digit can be any of the three choices, while the third digit has to be either 3 or 5) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
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Given Solution:: ** Using the box method, where in this case we draw three boxes and put one number in each: The 1st box can be filled with any of the three so the first number of possibilities is 3 The 2nd box can also be filled with any of the three so the second number of possibilities is 3 The last digit must be odd, so there are only 2 choices for the third box. By the Fundamental Counting Principle we therefore have 3*3*2=18 possible combinations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): question 11.2.56 (10th edition 50) 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups ********************************************* Your Solution: 10*4*6*3= 720 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
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Given Solution:: A setup consists of a guitar, a case, an amp and a processor. There are 10 choices for the guitar, 4 choices for the case, 6 choices for the amp and 3 choices for the processor. So by the Fundamental Counting Principle there are 10 * 4 * 6 * 3 = 720 possible setups. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok You may add comments on any surprises or insights you experienced as a result of this assignment: I just need to continue to be more careful when reading the questions and figuring out the solution. Most of the time I miss an entire problem because of one small mistake.