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# In how many ways can Ann, Bea, Cam, Don, Ella and Fey be

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Intern
Joined: 05 Dec 2008
Posts: 20
In how many ways can Ann, Bea, Cam, Don, Ella and Fey be [#permalink]

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06 Dec 2008, 19:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated in a row if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720
Intern
Joined: 05 Dec 2008
Posts: 20
Re: Counting challenge question #1 (easiest) [#permalink]

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07 Dec 2008, 14:24
Exactly - the total number of arrangements minus the number of arrangements where Ann and Bea are seated together.
Manager
Joined: 09 Jul 2008
Posts: 110
Location: Dallas, TX
Schools: McCombs 2011
Re: Counting challenge question #1 (easiest) [#permalink]

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07 Dec 2008, 17:11
Should it not be 600? 6!-5!.
If we consider both Ann and Bea as one atom, the number of combinations they can be together is 5!
Intern
Joined: 05 Dec 2008
Posts: 20
Re: Counting challenge question #1 (easiest) [#permalink]

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07 Dec 2008, 17:15
Not exactly.
Yes, it's a good idea to count both Ann and Bea as one element/atom (AB), in which case there are 5! arrangements.
But we also need to count both Bea and Ann as one element/atom (BA), in which case there are 5! arrangements as well.
In other words, there are two ways to arrange the single element/atom consisting of Ann and Bea.
Re: Counting challenge question #1 (easiest)   [#permalink] 07 Dec 2008, 17:15
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# In how many ways can Ann, Bea, Cam, Don, Ella and Fey be

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