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?

Part A

The question asks: 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?

So I am inferring that the 4 smarties must sit so that I can't sit between two smarties.

Scheme 1: All four smarties could position themselves in the middlemost grids--2 by 2 adjoining grids in the middle. For every sit one smartie takes, 9 are eliminated for the upcoming smartie--that is, the smartie's own sit and 8 others that prevent possible middle positions. That is, when smartie 1 sits, you move one grid off immediately surrounding grids, and eliminate sits that can allows for a middle smartie. Do this for other 3 smarties so that they all sit in the 2 by 2 middlemost grids. When this is done, I have 20 sits which I can take so that I am not in the middle of two smarties. Again, the there are 4! ways the middlemost smarties could sit if we are to allow for rotation and reflections.

Therefore number of possible sittings in this scheme is

4!*36*27*18*9*20 =

Scheme 2: The four smarties could position themselves at all corners so that immediately adjacent sits are eliminated. So this method yields 4!*36*32*28*24*20 = 18,579,456

We pick the larger number, which is from scheme 2

Part B:

We continue from Scheme 2 and note that if 5 smarties could sit in 5!*36*32*28*24*20 without violating the constraint, I have 19 ways of taking my sits when 5 smarties are seated so that the constraint isn't violated.

So we multiply 5!*36*32*28*24*20*19 to yield 1,765,048,320 ways.

The number of ways look like alot!

Posted by Osi on 2004-11-06 09:24:57