Count all possible paths from top left to bottom right of a mXn matrix


https://www.geeksforgeeks.org/count-possible-paths-top-left-bottom-right-nxm-matrix/
The problem is to count all the possible paths from top left to bottom right of a mXn matrix with the constraints that from each cell you can either move only to right or down
Examples :
Input :  m = 2, n = 2;
Output : 2
There are two paths
(0, 0) -> (0, 1) -> (1, 1)
(0, 0) -> (1, 0) -> (1, 1)
Note the count can also be calculated using the formula (m-1 + n-1)!/(m-1)!(n-1)!.
Another Approach:(Using combinatorics) In this approach We have to calculate m+n-2 C n-1 here which will be (m+n-2)! / (n-1)! (m-1)!
int numberOfPaths(int m, int n) {
            // We have to calculate m+n-2 C n-1 here 
            // which will be (m+n-2)! / (n-1)! (m-1)! 
            int path = 1;
            for (int i = n; i < (m + n - 1); i++) {
                path *= i;
                path /= (i - n + 1);
            }
            return path;
        }



    static int numberOfPaths(int m, int n)
    {
        // Create a 1D array to store results of subproblems
        int[] dp = new int[n];
        dp[0] = 1;
  
        for (int i = 0; i < m; i++) {
          for (int j = 1; j < n; j++) {
            dp[j] += dp[j - 1];
          }
        }
  
        return dp[n - 1];
    }



    //  the topmost leftmost cell (cell at 1, 1)
    static int numberOfPaths(int m, int n)
    {
        // Create a 2D table to store results 
        // of subproblems
        int count[][] = new int[m][n];
   
        // Count of paths to reach any cell in 
        // first column is 1
        for (int i = 0; i < m; i++)
            count[i][0] = 1;
   
        // Count of paths to reach any cell in
        // first column is 1
        for (int j = 0; j < n; j++)
            count[0][j] = 1;
   
        // Calculate count of paths for other 
        // cells in bottom-up manner using
        // the recursive solution
        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
   
            // By uncommenting the last part the 
            // code calculatest he total possible paths 
            // if the diagonal Movements are allowed
            count[i][j] = count[i-1][j] + count[i][j-1]; //+ count[i-1][j-1];
   
        }
        return count[m-1][n-1];
    }



    static int  numberOfPaths(int m, int n)
    {
        // If either given row number is first or 
        // given column number is first
        if (m == 1 || n == 1)
            return 1;
   
        // If diagonal movements are allowed then 
        // the last addition is required.
        return  numberOfPaths(m-1, n) + numberOfPaths(m, n-1);
                // + numberOfPaths(m-1,n-1);
    }






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