Advanced Math-Problem Solving

[kymtwoman]

I have no clue where to begin this problem:

You want to move a nickel from the lower left corner to the upper right corner of a 5 x 5 chessboard. You are only allowed to move the nickel one square at a time, and each move must be either to the right or up. How many different paths are possible?

I even started trying to draw out all of the possiblities, but got completely lost and confused. There has to be a mathematical way to solve this problem.

 

[mmm4444bot]

Try looking at the squares to which pieces could POSSIBLY move, each step of the way.

I'm thinking that it's the diagonals.

I mean, starting at the lower left corner, there are only 2 possible squares to which the piece could move in the first step.

If you subsequently move from either of those two squares, in the second step, there are three possible squares where you might end up.

Likewise, there are four possible squares on which to land for the third step.

So far, we have 2 * 3 * 4 possible paths.

(Notice that the 2, 3, and 4 squares are aligned diagonally.)

Continue reasoning, in this way.

I'm thinking that there are 4! * 5! possible paths to the upper right corner.

What do you think ?

 

[galactus]

The number of paths moving a nickel from the lower left to the upper right square on a 5-by-5 chessboard is the same as moving from the lower left corner to the upper right corner on a 4-by-4 grid (Say, moving a dot at the lower left vertex/corner up to the upper right vertex/corner).

There are 8 moves.

How many different strings of size 2n are there, consisting of n Right's and n Up's?. There are \(\displaystyle \binom{2n}{n}\)

This problem is very similar to the one where you arrange letters in a word and have repeating letters. For the 4-by-4

arrangment you have a total of 8 letters (4 x's and 4 y's) but each of these is repeated (x=right and y=up), so that the number of ways they can

be arranged is

\(\displaystyle \frac{(2n)!}{(n!)^{2}}=\binom{2n}{n}\), where n=4.

 

[iambryanatkinson]

move only UP and RIGHT right?

the simple idea is that amount of unique path to a square is sum of amount of unique path of any square that can move to the square

start with 1 path to lower left square since this is starting point
all squares on lower side and left side also has 1 unique path

on any square add the number of path from leftside and downside give its number of path

do this for all of the squares in chessboard, the number of path to the top right square is 70, exactly same value of galactus' answer