If the allowable moves are : one step towards Right, and one step towards
bottom
Explanation for (m+n) C n: you may represent the path using a String
consisting of only R and D,
where R is a right move and D is a downward move. To go from (1,1) to (m,n)
the length of the string
will be (m+n) as there are m rows so m 'D's' and n columns so n 'R's'. Now
the problem is to find the
number of strings of size m+n of only Rs and Ds such that it has exactly m
Ds and n Rs. So the answer is
(m+n) C n.


On Thu, Feb 21, 2013 at 2:14 PM, kumar ankit <[email protected]> wrote:

> Well, (m+n) C (n) is the answer only in the case the allowable moves are :
> one step towards Right, and one step towards bottom.
>
>
>
> On Thu, Feb 21, 2013 at 2:05 PM, shady <[email protected]> wrote:
>
>> How did you directly arrive at that solution ? Can you please explain
>>
>>
>> On Thu, Feb 21, 2013 at 1:52 PM, Gaurav Rana <[email protected]>wrote:
>>
>>> (m+n)C(n)
>>>
>>>
>>> On Thu, Feb 21, 2013 at 1:26 PM, shady <[email protected]> wrote:
>>>
>>>> Given a matrix of size mXn, find the number of paths from the top left
>>>> cell to the bottom right cell.
>>>>
>>>> BFS is one way... any other approach ?
>>>>
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>
>
>
> --
> Kumar Ankit
> Senior Undergraduate
> Department of Computer Engineering
> Institute of Technology
> Banaras Hindu University
> Varanasi
> Ph: +91 9473629892
>



-- 
Kumar Ankit
Senior Undergraduate
Department of Computer Engineering
Institute of Technology
Banaras Hindu University
Varanasi
Ph: +91 9473629892

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