# Mice running through grids

Ronald the Mouse stands on the bottom leftmost corner on a $$n\times n \text{ units}^2$$ grid. He can only move 1 unit up or 1 unit right at a time, prove that the number of paths Ronald the Mouse can take to get to the upper right corner will always be in the form $$\binom {2n}n$$ for any $$n \times n$$ grid

Note by Yan Yau Cheng
4 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

In order to move from the bottom left most corner to the upper right corner , Mr Mouse ( :P ) has to move n times upwards and n times to the right. Hence total no of moves is (n + n)!. Now this can be grouped into 2 groups of n moves of one type. Hence (n + n)!/n!*n!.

- 4 years, 1 month ago