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Permutations problem

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Permutations problem [#permalink]

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New post 10 Oct 2011, 10:20
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Hi there everyone,

Was wondering if someone could help me with a permutations question.

Six children, Arya, Betsy, Chen, Daniel, Emily and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A) 240
B) 480
C) 540
D) 720
E) 840

I figured that if it wasn't for the condition, the answer would be 6!. There are 10 ways Besty and Emily could be next to each other (5 ways BE, and 5 ways EB). So 6!/10=72 but it's not one of the answers.

The other way I can think of solving it would be to do 6! - 2(5!) which gives 480, a possible option. My reasoning here would be 6! is the total options, minus the number of permutations if we clumped EB or BE together.


If anyone can make this a bit clearer to me it would be very greatly appreciated!

Cheers,
Francis


Cheers,
Francis

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Joined: 20 Dec 2010
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Re: Permutations problem [#permalink]

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New post 10 Oct 2011, 11:39
packeted wrote:
Hi there everyone,

Was wondering if someone could help me with a permutations question.

Six children, Arya, Betsy, Chen, Daniel, Emily and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A) 240
B) 480
C) 540
D) 720
E) 840

I figured that if it wasn't for the condition, the answer would be 6!. There are 10 ways Besty and Emily could be next to each other (5 ways BE, and 5 ways EB). So 6!/10=72 but it's not one of the answers.

The other way I can think of solving it would be to do 6! - 2(5!) which gives 480, a possible option. My reasoning here would be 6! is the total options, minus the number of permutations if we clumped EB or BE together.


If anyone can make this a bit clearer to me it would be very greatly appreciated!

Cheers,
Francis


Cheers,
Francis


You have posted the question in the wrong forum. It should have been posted in the PS sub-forum.

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Re: Permutations problem [#permalink]

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New post 23 Oct 2017, 02:51
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Re: Permutations problem [#permalink]

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New post 23 Oct 2017, 03:38
packeted wrote:
Hi there everyone,

Was wondering if someone could help me with a permutations question.

Six children, Arya, Betsy, Chen, Daniel, Emily and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A) 240
B) 480
C) 540
D) 720
E) 840

I figured that if it wasn't for the condition, the answer would be 6!. There are 10 ways Besty and Emily could be next to each other (5 ways BE, and 5 ways EB). So 6!/10=72 but it's not one of the answers.

The other way I can think of solving it would be to do 6! - 2(5!) which gives 480, a possible option. My reasoning here would be 6! is the total options, minus the number of permutations if we clumped EB or BE together.


If anyone can make this a bit clearer to me it would be very greatly appreciated!

Cheers,
Francis


Cheers,
Francis


OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/six-children ... 10939.html
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Re: Permutations problem   [#permalink] 23 Oct 2017, 03:38
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