Tiling a Tilted Square

This week’s Fiddler is a challenging counting problem.

Consider the following array of 25 squares:

You are filling the array with rectangles by repeating the following two steps:

  1. Select one of the 12 squares along the outer perimeter that has not yet been selected as part of a rectangle.
  2. Form the largest rectangle you can that includes the square you just selected and other squares that are not yet part of any such rectangle.

You repeat these steps until every square along the perimeter has been selected. Here are two final states you might encounter:

How many distinct final states are possible? (Note: States that are rotations or reflections of each other should be counted as distinct.)

My solution:
[Show Solution]

The Likeliest Monopoly Square

This week’s Fiddler is about rolling dice in the board game Monopoly.

We have a square board with 40 individual spaces around it, numbered from 0 to 39. All players begin on space 0 (akin to the “Go” square in Monopoly) and roll a pair of dice to determine how many spaces they advance each turn. However, unlike Monopoly, there is no way to otherwise advance around the board (i.e., there’s no “Chance,” “Community Chest,” going to jail, etc.). In their first pass around the board, which space from 1 to 39 are players most likely to land on at some point (i.e., not necessarily on their first or last roll, but after any number of rolls)?

Extra Credit
The square board has 10 spaces on each side. The first side has spaces 0 through 9, the second side has spaces 10 through 19, the third side has spaces 20 through 29, and the fourth side has spaces 30 through 39. Because you’re rolling two dice, it’s impossible to land on space 1 in your first pass around the board. Several other spaces on the first side of the board are similarly unlikely. Putting that first side of the board aside, which space from 10 to 39 are players least likely to land on at some point during their first pass around the board? (Another question: What if you rolled three dice at a time instead of two?)

My solution:
[Show Solution]