This is the second in my series of posts in combinatorics. The first post links the Fundamental Counting Principle, Powers of 2, and the Pascal Triangle. This second post connects the Pascal’s Triangle and the formula for counting the number of permutations with identical objects. The context for connections is a puzzle about counting the total routes of a rook to squares in a chessboard.

###### The Puzzle

Trace the shortest route the rook can traverse from its corner position to the opposite corner. How many such routes are there?

Solution 1

A useful problem solving strategy here is to simplify the problem: Count the number of distinct paths the rook will land to the nearest square.

To determine the shortest path, the rook should either go north or east or left or right only. That is, the rook has only two choices each time it moves through a square as shown in the figure below. You will not reach the third or fourth line before you figure out that the pattern generates the Pascal’s Triangle.

Thus there are 3432 distinct shortest paths the rook can traverse from one corner to the opposite corner.

Solution 2

Another way to solve puzzle 2 is shown below. The arrows trace some possible shortest paths for the rook.

The paths of the arrows in the figure can be represented this way: (N = North, E = East)

- N N N E N E E N E N E N E E
- E E E E N N N E E N N N N E
- E E E E E E E N N N N N N N

Pascal’s Triangle is almost ‘magical’ in all of the mathematical connections it can be applied to. Thanks for sharing a ‘simple’ one. (And here I use the word, ‘magical’ lightly, and the word ‘simple’ as meaning ‘elegant.’

I have to be careful when introducing Pascal’s Triangle to my 6th graders, as I fear we’d never get back to the curriculum at hand.

And this might be a time to remind readers of “The Number Devil: A Mathematical Adventure.”