Assignment03_OROUR - 3.31

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3.31 a) Out of the possible drug combinations, there are fifteen different 2 drug combinations of the  six drugs possible. 6! / (2! * 4!) = 6*5*4*3*2*1 / (2*1) * (4*3*2*1) = 720 / 2*24 = 720/48 = 15 b) Out of the possible drug combinations, there are different twenty different 3 drug  combinations of the six drugs possible. (6!)/(3!*3!) = 6*5*4*3*2*1/(3*2*1)*(3*2*1) = 720 / 6*6 =  720/36 = 20 c) Out of the possible drug combinations, there are fifteen different 4 drug combinations of the  six drugs possible. 6! / (2! * 4!) = 6*5*4*3*2*1 / (2*1) * (4*3*2*1) = 720 / 2*24 = 720/48 = 15 d) Out of the possible drug combinations, there are six different 5 drug combinations of the six  drugs possible. 6! / (5! * 1!) = 6*5*4*3*2*1 / (5*4*3*2*1) * (1) = 720 / 120 = 6 e) I am only able to find 63 ways the drugs can be combined for this study. As listed above,  there are 56 multiple drug combinations. There are 6 single drug combinations, plus the control  treatment of no drugs. That's 56+6+1 = 63. I don't understand where a 64th drug combination  can be found. 3.42 a) The sample points for event A are 1-6, 2-5, and 3-4. The sample points for event B are 4-1,  4-2, 4-3, 4-4, 4-5, and 4-6. The sample points for the intersection of events A and B is 4-3. The 
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This note was uploaded on 07/28/2011 for the course STATISTICS BA 240 taught by Professor Winnieli during the Spring '11 term at Bellevue College.

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Assignment03_OROUR - 3.31

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