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Four Way Smarties (Posted on 2004-11-05) Difficulty: 3 of 5
You and four other people (who coincendentally are all smarties) are in late testing room where you will take your test where there is a 6 by 6 grid of equally spaced desks with chairs in the same relative spot.

You go into the room after all four smarties have chosen their location. You have a test taking policy where you always want to sit at the midpoint between two smarties. The smarties in the room with you feel the exact opposite way, so their arrangement is always such that no smartie is at the midpoint of two other smarties

However, depending on where the smarties are sitting, you may not be able to sit at the midpoint since in all cases it would always be where there is no chair and desk. (There is a strict no moving desks or chairs rule too.) How many ways could the current 4 smarties sit such that you couldn't sit at the midpoint of two smarties if reflections and rotations count as well?

How many ways could you not find where you want to sit if there were 5 smarties other than you and reflections and rotations count as well?

See The Solution Submitted by Gamer    
Rating: 4.2000 (5 votes)

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re(2): My new solution (Takes from previous solution) | Comment 17 of 22 |
(In reply to re: My new solution (Takes from previous solution) by Tristan)

let's examine this. Note that the smartie eliminates 4 seats when seated at the corners as shown below. There are other schemes where he can eliminate 9, but this gives less number of ways. The scheme that eliminates 4 seats is what creates the MAXIMUM number of ways the smarties and myself can sit. Here "E= empty, M = Myself". Note that positions 1 to 4 can be changed in 4! ways for 4 smarties and 5! ways for 5 smarties. Since I sit last, my position is dependent on the number of sits left so that I don't violate the constraint. 

1   e  =  =  e  2

e   e =  =   e  e

=  = =  =   = =

5  M =  =   = =

e  e =  =   e  e

3  e =   =  e  4

"9 seats lost" case

*1e   2e*    1e*   2e*   1e*   2e*

*3e   4e      3e    4e     3e     4e* 

*1e   2e      1      2      1e     2e*

*3e   4e      3      4      3e     4e*

*1e   2e      1e    2e    1e     2e*

*3e    *4e   *3e   *4e  *3e   4e*

In the above note that all sits appear to be eliminated (denoted with sit number and e=empty). However, there are 20 more sits that allow sittings without violating the question's constraint(*on the periphery). But this scheme produces less number of ways, so it makes sense to pick the larger number.


  Posted by Osi on 2004-11-06 12:10:03
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