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Solutions from a student (dragged) 30

# Solutions from a student (dragged) 30 - F must be a...

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possible choices. Part (b): If A and B will not searve together we can construct the total number of choices by considering clubs consisting of instances with A included but no B , B included by no A , and finally neither A or B included. This can be represented as 1 · 8 2 + 1 · 8 2 + · 8 3 = 112 . This result needs to again be multipled by 3! as in Part (a) of this problem. When we do so we find we obtain 672. Part (c): In the same way as in Part (b) of this problem lets count first the number of clubs with C and D in them and second the number of clubs without C and D in them. This number is 8 1 + 8 3 = 64 . Again multiplying by 3! we find a total number of 3! · 64 = 384 clubs. Part (d): For E to be an o ffi cer means that E must be selected as a club member. The number of other members that can be selected is given by 9 2 = 36. Again multiplying this by 3! gives a total of 216 clubs. Part (e):
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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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