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how many ways can 7 children be arranged in a circular table [#permalink]
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12 Feb 2005, 03:36
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how many ways can 7 children be arranged in a circular table where 2 chidren must be apart.



Manager
Joined: 13 Oct 2004
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(# 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!.



VP
Joined: 30 Sep 2004
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Location: Germany

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



SVP
Joined: 03 Jan 2005
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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.



VP
Joined: 13 Jun 2004
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Location: London, UK
Schools: Tuck'08

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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Joined: 30 Sep 2004
Posts: 1480
Location: Germany

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.



VP
Joined: 13 Jun 2004
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Location: London, UK
Schools: Tuck'08

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



VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

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!



VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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.



Director
Joined: 05 Jan 2005
Posts: 557

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 AG. 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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Joined: 30 Sep 2004
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Location: Germany

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 AG. 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 (n1)! instead of n! for the total ways of arrangements.



Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

I completely agree with prep_gmat's soln.
it will be !6  (2*!5)



Manager
Joined: 25 Oct 2004
Posts: 247

One more for 6! 5!2



Director
Joined: 27 Dec 2004
Posts: 898

Please explain 2 * 5!
I don't get it.



VP
Joined: 25 Nov 2004
Posts: 1483

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:
xabcdefy or yabcdefx.
therefore, the number of ways the children can be arranged: 2(5!).
=6!2(5!)=480










