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Circular probability

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by tdkk123 » Mon Sep 19, 2011 10:10 am
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?

OA [spoiler]1/60[/spoiler]


I have no clue where to start?

Also are problems like these common on the gmat? is it at only a 750+ level that you would get these problems are are they common on the lower levels as well. Im writing my gmat in a few days and problems like these are really throwing me way off
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Source: — Problem Solving |

by gmatclubmember » Mon Sep 19, 2011 10:20 am
n people can be arranged in (n-1)! ways on a circular arrangement. So total no. of ways/possibilities of seating are : 5!=120.
All the person can be seated either clockwise - ABCDEF or anticlockwise FEDCBA. So there are 2 favorable arrangements.
Probability = 2/120=1/60.
Hope it make it clear for you... :)
All the best for ur GMAT.

Cheers
Ami/-
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by shankar.ashwin » Mon Sep 19, 2011 10:23 am
For circular arrangements, the number of possible arrangements is given by (n-1)!

In this case; 6 people so (6-1)! = 120.

The arrangement mentioned in the problem is possible 2 ways (either clockwise or anti-clockwise) Draw the order to be clear

Hence; Prob of given order = 2/120 = 1/60.

I havnt seen a lot of circular arrangements in the GMAT, its not common in my opinion
tdkk123 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?

OA [spoiler]1/60[/spoiler]


I have no clue where to start?

Also are problems like these common on the gmat? is it at only a 750+ level that you would get these problems are are they common on the lower levels as well. Im writing my gmat in a few days and problems like these are really throwing me way off
Join the discussion

by cans » Mon Sep 19, 2011 10:23 am
well arrangement will be: FEDCBAF (I wrote F twice to show its circular..)
There are 2 arrangements: (anti clockwise or clockwise)
total =(6-1)! (for circular..)
thus prob = 2/120=1/60
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Cans!!
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