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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?

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?

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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This post received KUDOS

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
_________________

......................................................................... +1 Kudos please, if you like my post

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

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

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 anti-clockwise direction. So total 12 ways. Total ways is (6-1) ! = 120. So Probability = 12/120 = 1/10

gmatclubot

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

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