#$&* course MTH 152 2/18 7:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution:: ** C(9, 6) occurs in the n=9 row, the r=6 position of Pascal’s Triangle, which is the 10th row and the 7th number in the row. C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. (Note that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left.) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK question 11.4.18 clueless check of four of nine possible classrooms. How many of the possible selections will fail to locate the classroom? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: C(9,4) giving you 126 possible choices Fail to locate: C(8,4) giving you 70 possible choices to fail to locate classroom confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** There are C(9,4) possible combinations of the four classrooms to be checked. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check of 4 rooms to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. So there are 126 possible combinations of the four classrooms to be checked, and 70 of these combinations fail to include the ‘right’ classroom. Thus only 126 - 70 = 56 of the possible combinations will include the ‘right’ classroom. So the chance of finding the right classroom is 56 / 126, a little less than 50%. STUDENT QUESTION Can you work out how C (9,4) = 126 and how C (8,4) = 70??? Where did I go wrong??? I worked them out the way I thought I should. Am I getting combinations confused with permutations, or do I just have it totally wrong??? INSTRUCTOR RESPONSE I think you have mixed up permutations and combinations. You calculuated 9! / (9-4)!.. To clarify: C(9, 4) = 9! / ( 4! * (9-4)! ), not 9! / (9-4)!. 9! / (9-4)! = 9 * 8 * 7 * 6, which is the number of possible ordered choices of 4 objects chosen from among 9. This is the permutation. To get the number of unordered choices, you would divide the number of ordered choices by 4! (since every set of 4 choices could have appeared in any of 4! orders). So instead of 9! / (9-4)!., you would divide this by 4! to get 9! / ( 4! (9-4)! ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK