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# how many ways can 7 children be arranged in a circular table

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Director
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how many ways can 7 children be arranged in a circular table [#permalink]

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12 Feb 2005, 02:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

how many ways can 7 children be arranged in a circular table where 2 chidren must be apart.

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Manager
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12 Feb 2005, 08:47
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

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VP
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13 Feb 2005, 11:01
prep_gmat wrote:
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

do you multiply by 2 because it is a permutation problem ? otherwise pls explain ! thx

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SVP
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13 Feb 2005, 19:41
There are two ways to arrange the two boys who sit together, one way is for A to sit left to B, the other is for A to sit right to B.

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VP
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13 Feb 2005, 22:04
prep_gmat wrote:
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

Is this the OA ?
Please explain how did you get 2.5!

Thanks

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VP
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13 Feb 2005, 23:29
Antmavel wrote:
prep_gmat wrote:
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

Is this the OA ?
Please explain how did you get 2.5!

Thanks

take for example a and b as the two children. they sit in the first and second position. next to them there are 5*4*3*1 possible ways to sit the other children. we multiply by 2 because there are even more ways when a and b switch their seats.

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14 Feb 2005, 00:43
sorry christoph, I am still lost on this one I can't figure out the reasonning process

first, can you confirm if there are :

5*4*3*1 possible ways to sit the other children (like you've written) ?
or 5*4*3*2*1 possible ways to sit the other children ?

I am not sure I could answer right if I see this kind of questions again at GMAT test

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VP
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14 Feb 2005, 00:49
Antmavel wrote:
sorry christoph, I am still lost on this one I can't figure out the reasonning process

first, can you confirm if there are :

5*4*3*1 possible ways to sit the other children (like you've written) ?
or 5*4*3*2*1 possible ways to sit the other children ?

I am not sure I could answer right if I see this kind of questions again at GMAT test

oh a typo ! yes 5*4*3*2*1

first scenario: a b 5*4*3*2*1 = 5!
second scenario: b a 5*4*3*2*1 = 5!
=> 2*5!

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VP
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14 Feb 2005, 01:15
i got it

i read 2.5! (like 2,5) whereas your solution was 2*5! (2 multiplied by 5!)

sorry, now i totally understand it. Anyway, thank you for your time and for your explanation christoph, it was nice.

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Director
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14 Feb 2005, 04:01
christoph wrote:
Antmavel wrote:
prep_gmat wrote:
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

Is this the OA ?
Please explain how did you get 2.5!

Thanks

take for example a and b as the two children. they sit in the first and second position. next to them there are 5*4*3*1 possible ways to sit the other children. we multiply by 2 because there are even more ways when a and b switch their seats.

I'M completely lost here.

Let's say we have seats A-G. i.e. A,B,C,D,E,F,G
then, the no. of ways 2 people can stay together is:

AB, BC, CD, DE, EF, FG, GA (7.2 = 14 ways!)

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VP
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14 Feb 2005, 05:05
Arsene_Wenger wrote:
christoph wrote:
Antmavel wrote:
prep_gmat wrote:
(# of ways 7 children can be arranged around a circular table) - (# of ways the 2 children can be seated together)

# of ways 7 can be arranged in a circular pattern = 6!
# of ways 2 can be seated together = 2.5!

# of ways 2 children cannot be seated together in a circular pattern = 6! - 2.5!.

Is this the OA ?
Please explain how did you get 2.5!

Thanks

take for example a and b as the two children. they sit in the first and second position. next to them there are 5*4*3*1 possible ways to sit the other children. we multiply by 2 because there are even more ways when a and b switch their seats.

I'M completely lost here.

Let's say we have seats A-G. i.e. A,B,C,D,E,F,G
then, the no. of ways 2 people can stay together is:

AB, BC, CD, DE, EF, FG, GA (7.2 = 14 ways!)

this is a problem of circular permutation. imagine a circle with 3 seats A B C:

first arrangement is A B C
second is A C B

that is because the circle can be rotated. that is why we use (n-1)! instead of n! for the total ways of arrangements.

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14 Feb 2005, 07:09
I completely agree with prep_gmat's soln.

it will be !6 - (2*!5)

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Manager
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14 Feb 2005, 15:41
One more for 6!- 5!2

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Director
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15 Feb 2005, 11:18
I don't get it.

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VP
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15 Feb 2005, 22:50
Folaa3 wrote:
Please explain 2 * 5! I don't get it.

SUPPOSE THE TWO CHILDREN ARE X AND Y AND OTHERS ARE A,B,C,D, AND E.

The following positin is the way two children seat togather:
x-a-b-c-d-e-f-y or y-a-b-c-d-e-f-x.

therefore, the number of ways the children can be arranged: 2(5!).

=6!-2(5!)=480

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15 Feb 2005, 22:50
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# how many ways can 7 children be arranged in a circular table

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