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There are 5 white plates and 1 blue plate in a circle, with

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There are 5 white plates and 1 blue plate in a circle, with [#permalink] New post 13 Mar 2008, 16:34
There are 5 white plates and 1 blue plate in a circle, with only one of 6 different foods in each plate. Among the foods, there are peanuts and raisins, how many arrangements are there if the blue plate only contains peanuts or raisins?
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Re: Combination problem [#permalink] New post 13 Mar 2008, 17:22
240

With Peanuts in blue plate remaining foods can be arranged in 5! ways. Similary with raisins in blue plate too. so total number of combinations is 5!+5! = 240. If the plates are not circular the answer would be different because we don't need to keep them in the order.
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Re: Combination problem [#permalink] New post 13 Mar 2008, 17:29
I get it..thanks!!!
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Re: Combination problem [#permalink] New post 13 Mar 2008, 21:06
what happen if they were circular, though?
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Re: Combination problem [#permalink] New post 13 Mar 2008, 21:32
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In case we don't need to care about the order, it comes down to just 2 arrangements.
Re: Combination problem   [#permalink] 13 Mar 2008, 21:32
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There are 5 white plates and 1 blue plate in a circle, with

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