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counting problem

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Junior | Next Rank: 30 Posts
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by sidon » Mon Apr 12, 2010 10:58 am
Need help with a counting problem ...

In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be
seated next to each other?
(A) 240
(B) 360
(C) 480
(D) 600
(E) 720

Thanks,
Sidon
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by akahuja143 » Mon Apr 12, 2010 11:22 am
IMO B.. 6!/2! -- =360
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by November Rain » Mon Apr 12, 2010 11:36 am
Hi
I think the answer is C

The best way to way to solve this question is to use a method i learned with Manhattan.

Instead of trying to figure out the possible combinations of having both girls separated by one or more people, try figure out the possible combinations of having them together, and then subtract if from the total number of possible combinations.

So, the possible combinations is 6! = 720

On the other hand there are 5! * 2 = 240 possible combinations of both girls being together:
- You put one girl "glued" to another as if the two girls were just one, and you will have 5! possible combinations
- Then you multiply by 2, because you also need to count the possibilites of the two girls that you glued together switch places.


Finally you subtract the 240 to the 720, and you will get 480.

Could you confirm the OA?
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by sidon » Mon Apr 12, 2010 11:48 am
Thanks - correct answer is C.

Thanks for answeing - it is a good way for solving the problem ...
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by akahuja143 » Mon Apr 12, 2010 12:06 pm
Thanks for explanation November Rain sounds like a good way to approach the problem like this
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