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Question

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

Anurag@Gurome · Answer

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 )

anshumishra · Answer

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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