Unique Paths in 2D grid

Unique Paths 62 - LeetCode
There is an m x n grid. One can only move either down or right at any point in time. One is trying to reach the bottom-right corner of the grid.
How many possible unique paths are there?
Paths[i][j] = paths[i][j-1] + paths[i-1][j]
int no_of_paths(int width, int height)
{
 int i,j;
 int* no = new int[width];

 for (i=0 ;i< width; i++)
 {
  no[i] = 1;
  cout << "1 "; //just for printing/debugging purpose
 }
 cout << endl;     //just for printing/debugging purpose

 for (i=1 ;i< height; i++)
 {
  cout << no[0] << " ";     //just for printing/debugging purpose
  for (j=1; j< width; j++)
  {
   no[j] = no[j]+no[j-1];
   cout << no[j] << " "; //just for printing/debugging purpose
  }
  cout << endl;             //just for printing/debugging purpose
 }
 int res = no[width-1];

 delete [] no;
 return res;
}
http://joaoff.com/2008/01/20/a-square-grid-path-problem/

1. all the paths have size 2×n (the reason is obvious: you have to go right n positions and down another n positions);
2. since we can only go right or down, we can identify every path by a string of Rs and Ds, where a R means going right and a D means going down; as an example, the paths illustrated in the problem statement are (from left to right and from top to bottom): RRDD, RDRD, RDDR, DRRD, DRDR and DDRR;
3. the strings mentioned above must contain the same number of Rs and Ds.

From these three observations, we can transform the problem to the following:

How many different strings of size 2×n, consisting of n Rs and n Ds, are there?

(2nn)=(2n)!n!×n!    .

Generalization to m×n grids

The generalization to an m×n grid is also simple. The only difference is that the strings have length m+n. Using the same reasoning as above, the number of paths through an m×n grid is:

(m+nn)=(m+n)!m!×n!  

(m+nn)=(m+n)!m!×n!    .

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