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

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TopNewestIn 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!.

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