Problem Solving

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



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Answer is 480
Solution: All 6 can occupy 6 places in 6! ways.
Now lets assume, Ann and Bea always sit together and consider them as one unit. So 5 people can occupy 5 places in 5! ways. Also Ann and Bea (whom we have considered one unit) can occupy the two places next to each other in 2 ways. Hence the total no. of ways when Ann and Bea sit together = 2*5!.
So, the no. of ways where Ann and Bea do not sit together =(total no of ways)- (no. of ways when they always sit together) = 6!-2*5! = 480

Here's the long form approach that I finally figured out. Takes too long on the test, but it helped me visualize the problem.

We have 6! ways of arranging the people, so that's our starting point -> 720

I've setup a chart like so - If we have A and B assigned, we have 4 x 3 x 2 x 1 options with the remaining seats. So each row has 24 arrangements that we'll need to exclude.

1 2 3 4 5 6
A B 4 3 2 1
B A 4 3 2 1
4 A B 3 2 1
4 B A 3 2 1
4 3 A B 2 1
4 3 B A 2 1
4 3 2 A B 1
4 3 2 B A 1
4 3 2 1 A B
4 3 2 1 B A

Since there are ten total rows * 24 exclusions per row, we get 240 exclusions.

Why can't we stick with the shortest method as suggested by Rahul? After all this much knowledge of Permutations is expected of all GMAT aspirants. On GMAT, a logic is less qualified if it's longer than an already existing logic known to all.

Thanks

Sometimes non-standard due to you - this is awfully helpful !