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Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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15 Apr 2012, 21:16

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Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

A. 360 B. 120 C. 80 D. 240 E. 60

Total # of arrangement of 6 people is 6!.

In half of the cases Susan will be seated left to Tim and in half of the cases Susan will be seated right to Tim (why should one seating arrangement have more ways to occur than another?).

So, # of arrangements to satisfy the given condition is 6!/2=360.

Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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27 Mar 2014, 12:26

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Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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12 Apr 2014, 01:19

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satsymbol wrote:

Why Can not I use Glue method here? SK together, with 4 others - 5! = 120 ways.

Hi,

When you are using the above method what you are assuming is that they are sitting next to each other always, which is not what the question states. The question only says that S is always sitting left of T, maybe next, maybe away 1 chair ...there is no constraint on that.

----------------------------------- Kudos, if the post helped

Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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12 Apr 2014, 19:27

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ind23 wrote:

satsymbol wrote:

Why Can not I use Glue method here? SK together, with 4 others - 5! = 120 ways.

Hi,

When you are using the above method what you are assuming is that they are sitting next to each other always, which is not what the question states. The question only says that S is always sitting left of T, maybe next, maybe away 1 chair ...there is no constraint on that.

----------------------------------- Kudos, if the post helped

Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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26 Apr 2015, 20:34

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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24 Jun 2015, 03:37

Bunuel wrote:

gmihir wrote:

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

A. 360 B. 120 C. 80 D. 240 E. 60

Total # of arrangement of 6 people is 6!.

In half of the cases Susan will be seated left to Tim and in half of the cases Susan will be seated right to Tim (why should one seating arrangement have more ways to occur than another?).

So, # of arrangements to satisfy the given condition is 6!/2=360.

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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24 Oct 2016, 10:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 [#permalink]

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25 Jun 2017, 10:58

reto wrote:

Bunuel wrote:

gmihir wrote:

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

A. 360 B. 120 C. 80 D. 240 E. 60

Total # of arrangement of 6 people is 6!.

In half of the cases Susan will be seated left to Tim and in half of the cases Susan will be seated right to Tim (why should one seating arrangement have more ways to occur than another?).

So, # of arrangements to satisfy the given condition is 6!/2=360.

If the question asked that Susan should always sit DIRECTLY left to TIM, then the total # of arrangements would be:

5! = 120 ?

Because you can "glue" Kim and Susan "together". But you still have 6 chairs, how would you account for that?

Further: anyone also understod this part "Susan is seated always left to Tim" in the way that Susan needs to be seated directly left to Tim?

Thanks

Hi reto,

The highlighted portion says that susan is sitting left to Tim, it doesn't necessarily mean next to him BUT always left to him. i.e. 1 way is S,T,_,_,_,_,_, next S,_,T,_,_,_,_ and so on.

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