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Unformatted text preview: F must be a president we have two cases. The Frst is where F serves and is the president and the second where F does not serve. When F is the president we have two permutations for the jobs of the other two selected members. When F does not serve, we have 3! = 6 possible permutions in assigning titles amoung the selected people. In total then we have 2 ± 9 2 ² + 6 ± 9 3 ² = 576 , possible clubs. Problem 4 (anwsering questions) She must select seven questions from ten, which can be done in ± 10 7 ² = 120 ways. If she must select at least three from the Frst Fve then she can choose to anwser three, four or all Fve of the questions. Counting each of these choices in tern, we Fnd that she has ± 5 3 ²± 5 4 ² + ± 5 4 ²± 5 3 ² + ± 5 5 ²± 5 2 ² = 110 . possible ways....
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.
 Winter '08
 G.JOGESHBABU
 Probability

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