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February 7th, 2014, 03:28 PM  #1 
Newbie Joined: Feb 2014 Posts: 2 Thanks: 0  probability problem in hexagon
So say I have 6 bugs standing on the 6 vertices of a hexagon, one per vertex. And say they each pick a vertex that they are not currently on, and starts moving in a straight line towards that vertex at the same speed. So my question is how many possibilities are there for the bugs to move to the vertices such that none of them are ever in the same place at the same time?

February 7th, 2014, 03:51 PM  #2 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 644 Thanks: 85  Re: probability problem in hexagon
This seems like it should simply be 6! = 720 (719 other than the starting position).

February 7th, 2014, 04:01 PM  #3 
Newbie Joined: Feb 2014 Posts: 2 Thanks: 0  Re: probability problem in hexagon
I don't think so. No lines can intersect with another line at a point that is the same distance away from both starting vertices, so I think there are less.

February 8th, 2014, 06:43 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: probability problem in hexagon Hello, leuler! Quote:
The answer seems to be the number of derangements of 6 objects: But you remind us that the bugs must not collide . . . ack!  

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