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Manager
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New post 31 May 2006, 13:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

no answer choices given but good practise for everyone I think

12 players out of which 8 have to select in two different rows out of which in one row 3 person fixed and in other row two fixed. How many arrangement can possible to make two row 4 in each side.
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New post 31 May 2006, 14:24
Selecting 5 people out of 12 for the 5 "fixed" seats = C(12, 5).
Selecting another 3 from 7 for the remaining 3 "movable seats" (permutations) = P(7, 3)

Thus answer = C(12, 5)*P(7, 3)
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  [#permalink] 31 May 2006, 14:24
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