#$&* course Mth 152 October 6, 2011 - 3:00pm question 11.2.12 find 10! / [ 4! (10-4)! ] without calculator
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Given Solution:: ** Starting with 10! / [ 4! * (10-4) ! ], we replace (10 - 4) by 6 to get 10! / ( 4! * 6! ). Writing out the factorials our expression becomes 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] . The numerator and denominator could be multiplied separatedly then divided out but it's easier and more instructive to divide out like terms. Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by the same expression ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves us 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1). Every factor of the denominator divides into a number in the numerator without remainder: • Divide 4 in the denominator into 8 in the numerator, • divide 3 in the denominator into 9 in the numerator and • divide 2 in the denominator into 10 in the numerator and you end up with • 5 * 3 * 2 * 7 = 210. NOTE ON WHAT NOT TO DO: You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divide 3628800 by 16480. But that process would lose accuracy and be ridiculously long for an expression like 100 ! / ( 30! * 70!). For this calculation the numbers involved would be hundreds of digits long. Much better to simply divide out like factors until the denominator goes away, as it always will with expressions of this form. (form n ! / (r ! * (n - r)! ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OKAY ------------------------------------------------ Self-critique Rating: OKAY question 11.2.31 (10th edition #25) 3 switches in a row; find count prin to find # of possible settings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 3 switches in a row - each switch can be either on or off. The first switch has two settings, so does the second and the third. 2*2*2* = 8 possible settings confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OKAY - I did figure out the difference between the 6 orders and the 8 possible settings. Yay! ------------------------------------------------ Self-critique Rating: OKAY question 11.2.31 (10th edition #27) If no two adjacent switches are off why does the fundamental counting principle not apply? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: You need independent variables. Since the switches affect each other they are not independent. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION I thought for sure I had this right, but apparently I am confused. A switch is considered independent when it is concerned with the placement of the other switches. It is dependent if it is not concerned. These terms seem backward to me. INSTRUCTOR RESPONSE The given condition is that no two adjacent switches can be off. So for example if the first switch is off, the second switch cannot also be off. Or if the middle switch is off, neither of the other two can be off. Thus in some circumstances the state of one switch has an influence on the state of the other switches, meaning that its state is dependent on the state of the other switch, and the Fundamental Counting Principle does not apply. I believe that in this case, you did simply reverse the meanings of 'dependent' and 'independent'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Confused a bit about independent and dependent now. Did I get this right? Self-critique Rating: 3 question 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We can write or draw out the options of this set. It’s easier to draw them in using a box but because we are in word I’m going to write them. For the first box, we can put in any of the numbers but the possibility will be 3. For the second box, it can be any of the three numbers so the possibility is 3 again. Since we are looking for an odd digit and we only have 2 odd numbers to end the boxes with the possibility is only 2. We write the counting formula as so= 3*3*2=18 possible 3 digit combinations. confidence rating #$&*: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): OKAY ------------------------------------------------ Self-critique Rating: OKAY question 11.2.56 (10th edition 50) 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There are 10 possible set ups for the guitars, 4 for the cases, 6 for the amps, and 3 for the processors. We set the counting formula up, like so: 10*4*6*3=720 combination setups. confidence rating #$&*: 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): OKAY You may add comments on any surprises or insights you experienced as a result of this assignment: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This set seemed easier. We looked more at the foundation formulas and principles instead of more complicated combinations. ------------------------------------------------ Self-critique Rating: OKAy " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!