BTG set, Counting and probability 700+

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Jan 03, 2011 1:08 am

In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

My solution is different from the official one, I need evaluation/confirmation
apply the restrictions first and follow with the non-restricted parameters: Ann can be seated in four ways, Bob can be seated in three ways, Chuck, Don and Edd in any order can be seated in 3, 2 and 1 ways (no restrictions here) => 4*3*3*2*1=72 ways

Official explanation is attached

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Source: Beat The GMAT — Problem Solving | Mon Jan 03, 2011 1:08 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Jan 03, 2011 1:19 am

Night reader wrote:In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

The OA you provided is a traditional one. Let me explain.

Total number of arrangements = 5! = 120
Now treat Ann and Bob as a single individual, i.e. the case they always seat next to each other. Number of possible arrangements = 4! = 24. But in between them they can rearrange themselves in 2! ways too. Hence total number of arrangements such that they always seat next to each other = 2*24 = 48

Therefore, total number of arrangements such that they never seat next to each other = total number of arrangements without any restriction - total number of arrangements such that they always seat next to each other = (120 - 48) = 72

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Tue Jan 04, 2011 8:26 am

Night reader wrote:In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

My solution is different from the official one, I need evaluation/confirmation
apply the restrictions first and follow with the non-restricted parameters: Ann can be seated in four ways, Bob can be seated in three ways, Chuck, Don and Edd in any order can be seated in 3, 2 and 1 ways (no restrictions here) => 4*3*3*2*1=72 ways

Official explanation is attached

I would normally solve it as the Official solution. Here is another way to look at the problem ( Check if this is how you have solved it)

1 2 3 4 5

Case 1: A is on extreme end
If A is placed either at 1st place or 5th place, B can be placed in 3 ways, and the rest 3 in 3! ways
So, total no. of ways in this case = 2*3*3! = 36

Case 2: A is in the middle (not end)
If A is placed at any position other than 1 and 5, then B can be placed 2 ways, and the rest 3 can be placed in 3! ways
So, total no. of ways in this case = 3*2*3! = 36

Hence total number of possible placements win which A and B are not placed side by side = 36 + 36 = 72

Thanks
Anshu

(Every mistake is a lesson learned )

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