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February 7th, 2014, 03:28 PM   #1
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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?
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February 7th, 2014, 03:51 PM   #2
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Re: probability problem in hexagon

This seems like it should simply be 6! = 720 (719 other than the starting position).
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February 7th, 2014, 04:01 PM   #3
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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.
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February 8th, 2014, 06:43 AM   #4
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Re: probability problem in hexagon

Hello, leuler!

Quote:
I have 6 bugs standing on the 6 vertices of a hexagon, one per vertex.
They each pick another vertex and start moving directly toward 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 [color=blue]none of them are ever in the same place at the same time[/color]?

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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