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Randomly, six people A, B, C, D, E, and F sit around a [#permalink]
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24 Sep 2006, 08:26
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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



Manager
Joined: 28 Aug 2006
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Location: Albuquerque, NM

No of ways to arrange 6 people in circular table is 5! an dfor n people is (n1)!
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



Manager
Joined: 28 Aug 2006
Posts: 243
Location: Albuquerque, NM

No of ways to arrange 6 people in circular table is 5! an dfor n people is (n1)!
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



Manager
Joined: 08 Jul 2006
Posts: 89

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.
___________________________________________________________
The total number of possible combinations (6 things taken 3 at a time)
My guess is 3/10. ?????



Senior Manager
Joined: 31 May 2006
Posts: 368
Location: Phoenix AZ

is it 1/60?
2 ways to arrange. And total number of arrangemnets = 5!
so 2/120



Intern
Joined: 18 Sep 2006
Posts: 43

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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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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink]
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31 Jul 2014, 16:13
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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]
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02 Aug 2014, 13:35
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



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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink]
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11 Jan 2015, 04:29
Since its random the first seat should be 1/6. After that, your order is given and the probability that to be selected is equal. The chance to sit the right person randomly besides the first person is 1/5. For the next is 1/4 etc.
Since we got a round table, we can also reverse each seating arrangement and still get the right partners for each person. > *2
Since we can start with 6 different people, there are 6 different ways > *6
Its (2*6)/(6!) > 1/(5*4*3)



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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink]
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12 Jan 2015, 11:36
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Total number of ways in arranging 6 ppl in a cricle = (61)! = 120
with the given constarints, we can have only 2 arrangements one with B to left of A and other with Right of A.
probabilty = 2/120 = 1/60



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Re: Randomly, six people A, B, C, D, E, and F sit around a [#permalink]
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30 Apr 2015, 01:10
There can be 6 ways for the arrangement in clockwise direction and 6 ways in anticlockwise direction. So total 12 ways. Total ways is (61) ! = 120. So Probability = 12/120 = 1/10




Re: Randomly, six people A, B, C, D, E, and F sit around a
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30 Apr 2015, 01:10








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