#$&* course Mth 152 1/24 1:27 PM question 11.2.12 find 10! / [ 4! (10-4)! ] without calculator
.............................................
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 denominatorgoes away, as it always will with expressions of this form. (form n ! / (r ! * (n r)! ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: question 11.2.31 (10th edition #25) 3 switches in a row; fund count prin to find # of possible settings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For 3 switches in a row, use FCP to find the number of possible settings so there are 2 possible settings for the 1st switch, two for the 2nd, and 2 for the 3rd. 2*2*2=8 possible settings for 3 switches. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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 isindependent 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): ------------------------------------------------ Self-critique Rating: 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: The reason does NOT apply because FCP requires that events be independent of each other and in this problem the state of one switches influences the of state of all the other switches. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
Given Solution:: ** The reason the principle doesnt apply is that the Fund. Counting Principle requires that the events be independent. Here the state ofone switch influences the state of its neighbors (neither neighbor can be the same as that switch). So the choices are not independent. The Fund. Counting Principle requires that the events be independent. Since this is not the case the principle does not apply. ** STUDENT QUESTION I thought for sure I had this right, but apparently I am confused. A switch is considered independent when it is concerned with the placement of the other switches. It is dependent if it is not concerned. These terms seem backward to me. INSTRUCTOR RESPONSE The given condition is that no two adjacent switches can be off. So for example if the first switch is off, the second switch cannot also be off. Or if the middle switch is off, neither of the other two can be off. Thus in some circumstances the state of one switch has an influence on the state of the other switches, meaning that its state is dependent on the state of the other switch, and the Fundamental Counting Principle does not apply. I believe that in this case, you did simply reverse the meanings of 'dependent' and 'independent'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: question 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The first box can be filled with any of the three so there are 3 possible, the 2nd can be filled with any of the 3 so the 2nd number of possibilities is 3 and the 3rd has to be odd so there are only 2 choices for that one and by using FCP (3*3*2) we find there are 18 possible combinations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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): ------------------------------------------------ Self-critique Rating: question 11.2.56 (10th edition 50) 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There are 10 choices for the guitar, 4 choices for the case, 6 choices for the amps, and 3 choices for the processors so by using FCP we do 10*4*6*3=720 possible setups. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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): You may add comments on any surprises or insights you experienced as a result of this assignment: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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): You may add comments on any surprises or insights you experienced as a result of this assignment: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #(*!