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Seating Arrangements!

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by varun7nurav » Sun Sep 18, 2011 8:57 am
In how many ways can A, B, C, D, E and F be seated if A and B cannot be seated next to each other?
(A)240
(B)360
(C)480
(D)600
(E)720

Please also suggest a good way to handle problems such as these, I always get a little confused in which is the right method to approach.

Thank you!
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by sl750 » Sun Sep 18, 2011 11:29 am
First consider the case without the restriction. We can arrange 6 people in 6! ways

Now this case includes the case where A and B are seated together and without

Let's consider the case where A and B are seated together. We will treat them as 1 unit. Therefore
we now have 5 people. These 5 people can be arranged in 5! ways, and within the one unit (AB), we have 2 arrangements.

Total arrangements without AB seated together = Total arrangements- Total arrangements with AB seated
together

6!-5!*2
5!(6-2) = 120*4 = 480
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by varun7nurav » Sun Sep 18, 2011 11:43 am
great! Thank u!
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