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?

(In reply to My new solution (Takes from previous solution) by Osi)

To tell the truth, I didn't really read all your solutions before making that comment.  Shame on me!  But I have read your latest solution, and I still think you made a few mistakes.

The biggest mistake made throughout your solution is thinking that there are some positions that eliminate less seats.  You seem to have forgotten that the smarties may sit all the way across the class and still have a desk at their midpoint.

------  In this chart, the Xs mark the positions
XMX---  of 3 smarties, and Ms mark desks that are
------  midpoints.
--MM--
------
----X-

So no matter where the smarties sit, the corner or center, they always eliminate 8 desks for other smarties.

Now for the more negligible errors.

Part A:
Scheme 1:

Now to look at your multiplication.
4!*36*27*18*9*20

First of all, if you assume the smarties sit in the center, then the first smarty does not have 36 desks to choose from, but rather 4.  But as I showed above, it doesn't matter whether they sit in the center or not.

Also, you factored in 4! twice, though the other 4! is hidden.  The multiplication could be written as 4!*4!*9*9*9*9*20.  In my solution, I didn't factor 4! in at all, but factoring it in once is still ok (just a slight interpretation difference).  Factoring it in twice does not work.  Remember when you factor 36, you are counting all combinations where smarty 1 could be in any of the 4 smarty locations.

You factored in a 20 for the number of seats "you" can sit in.  I think this is just a slight interpretation difference from what I did.

Scheme 2:

4!*36*32*28*24*20 = 18,579,456

Again, if you assume that the smarties sit in the corners, the first one has 4 seats to choose from, not 36.  Also, each smartie eliminates 8 other seats, not 4.  Again, the extra 4! is not needed.

Part B:

The same mistakes are made here.  5! doesn't need to be factored in.  Each smarty eliminates 8 seats, not 4.  Also, if you assume all smarties sit in corners, note that you are using 5 corners when there are only 4 available.

Posted by Tristan on 2004-11-06 10:30:26