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?

To solve this problem, I placed one smarty in the corner of a 6x6 grid of desks.  Then, I made Xs on all the spots where if another smarty had sat, there would be no desk at the midpoint between them.  This is how it turned out.

OX X X
XXXXXX
 X X X
XXXXXX
 X X X
XXXXXX

Using this chart, I concluded that in order for no desks at midpoints to exist, none of the smarties can be both an even number of desks to the left/right and front/back.  Therefore, if I divide the desks into four sets like shown in the chart below, the smarties each sit in a different set.

121212
343434
121212
343434
121212
343434

So, with 4 smarties, each must sit in a different numbered set.  Each set contains 9 of the 36 desks.  Therefore, the number of combinations is 9^4, or 6561.

With 5 smarties, it is obvious that there must be at least one of the 4 sets with more than 1 smartie, therefore the number of combinations is 0.

Posted by Tristan on 2004-11-05 19:23:28