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Randomly, six people A, B, C, D, E, and F sit around a

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Randomly, six people A, B, C, D, E, and F sit around a [#permalink] New post 24 Sep 2006, 07:26
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Randomly, six people A, B, C, D, E, and F sit around a circular table. What is the probability that A is in middle of B and F, B is in the middle of A and C, C is in the middle of B and D, D is in the middle of C and E, E is in the middle of D and F, F is in the middle of A and E?

not sure of the answer
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 [#permalink] New post 24 Sep 2006, 07:46
No of ways to arrange 6 people in circular table is 5! an dfor n people is (n-1)!

Thus for each case prabability is 3!/5!

Multiplying 6 times the answer is 1/20^6 but seems a little too low

but if the question involved or between different cases then answer is
6*(1/20)

What is the source and what is the answer
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 [#permalink] New post 24 Sep 2006, 07:48
No of ways to arrange 6 people in circular table is 5! an dfor n people is (n-1)!

Thus for each case prabability is 3!/5!

Multiplying 6 times the answer is 1/20^6 but seems a little too low

but if the question involved or between different cases then answer is
6*(1/20)

What is the source and what is the answer
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 [#permalink] New post 24 Sep 2006, 09:46
Not too sure here...

An arrangement for A between B and F would essentially be the same as BAF or FAB right?

Thus, I employed the combination counting technique.

The number of possible outcomes BAF, ABC, BCD, CDE, DEF , and AFE.
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The total number of possible combinations (6 things taken 3 at a time)

My guess is 3/10. ?????
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 [#permalink] New post 24 Sep 2006, 10:11
is it 1/60?

2 ways to arrange. And total number of arrangemnets = 5!

so 2/120
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 [#permalink] New post 24 Sep 2006, 10:34
Should be 1/60...Total no of arrangements=5!=120..tw possible arrangements that satisfy the condition- aniclockwise and clockwise..so 2/120=1/60...what's the answer? :?:
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 [#permalink] New post 24 Sep 2006, 10:55
it is GMAt question.. not sure about the answer
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 [#permalink] New post 24 Sep 2006, 11:42
What did you pick? OR Do you remember any of the choices at least...We're all curious now......

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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink] New post 31 Jul 2014, 15:13
rkatl wrote:
Randomly, six people A, B, C, D, E, and F sit around a circular table. What is the probability that A is in middle of B and F, B is in the middle of A and C, C is in the middle of B and D, D is in the middle of C and E, E is in the middle of D and F, F is in the middle of A and E?

not sure of the answer


There are 6! ways of assigning six people A,B,C,D,E, and F around the table.
There are two patterns that satisfy the requirement: E D C B A F or F A B C D E
Finally, there are 6 ways to assign each people in the seat : A can be seat number 1, 2,3,4,5 ,6. Once you put A in one seat, other people have to sit according to A's seat ( two patterns that we discussed above).
In total, there are 6*2 = 12 to satisfy the requirements.
So the answer is 12 / 6! = 1/60
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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink] New post 02 Aug 2014, 12:35
rkatl wrote:
Randomly, six people A, B, C, D, E, and F sit around a circular table. What is the probability that A is in middle of B and F, B is in the middle of A and C, C is in the middle of B and D, D is in the middle of C and E, E is in the middle of D and F, F is in the middle of A and E?

not sure of the answer


I think the answer is 1/60

no. of ways arranging 6 people around a table =6!/2! = 360
However, the order is already fixed for all the people so they can be rotated in different ways around the table
Therefore, ans is 6/360 = 1/60
Re: Randomly, six people A, B, C, D, E, and F sit around a   [#permalink] 02 Aug 2014, 12:35
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