cannot be seated next to each other

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krishnasty: Master | Next Rank: 500 Posts; Posts: 142; Joined: Mon Jan 10, 2011 8:03 am; Thanked: 19 times

by krishnasty » Fri Jul 22, 2011 10:42 pm

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) 7202.

OA : C

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Source: Beat The GMAT — Problem Solving | Fri Jul 22, 2011 10:42 pm

knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Fri Jul 22, 2011 11:56 pm

Lets first find the total # of ways they can be arranged without restrictions
6P6=6!=720.......(1)

Now, lets find the # of ways that Ann and Bea will always be together

Ann Bea Cam Don Ella Fey

Lets replace Ann and Bea with an X

X Cam Don Ella Fey can be arranged amongst themselves in 5P5 ways=5!=120....(a)
But, inside of X, Ann and Bea can be arranged in 2 ways ie. Ann Bea or Bea Ann...So 2 ways.

Total # of ways in which Ann and Bea will always be together is 120*2=240 ...(2)

Total # of ways they can be arranged with no restrictions-Total # of ways in which Ann and Bea will always be together=Total # of ways where Ann and Bea will never be together.

Total # of ways where Ann and Bea will never be together=720-240=480 hence C

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