Finding No. of paths in matrix

Problem : Given a matrix of size m*n of 1’s and 0’s find the Number of paths from  matrix[0][0] to matrix[m-1][n-1], using the following moves :

if a[i][j] == 1 , you can move right and downwards.

if a[i][j] == 0 , move not allowed.

For ex :

1    1     1   1

1    1     1   1

1    0      1   1

1     1      1   1

No of paths = 11

Solution :

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class NoOfPathsInMatrix {

	int m,n;
	int arr[][]=null;
	
	public static void main(String[] args)throws IOException {

		NoOfPathsInMatrix obj=new NoOfPathsInMatrix();
		
		BufferedReader br =new BufferedReader(new InputStreamReader((System.in)));
		
		System.out.println("Enter the no of rows in matrix : ");
		obj.m = Integer.parseInt(br.readLine());
		System.out.println("Enter the no of cols in matrix : ");
		obj.n = Integer.parseInt(br.readLine());
		
		obj.arr = new int[obj.m][obj.n];
		for(int i=0; i< obj.m;i++)
		{
			System.out.println("Enter "+(i+1)+"th row : ");
			for(int j=0;j<obj.n;j++)
			{
				String s = br.readLine();
				obj.arr[i][j] = Integer.parseInt(s);
			}
		}
		
		System.out.println("No of paths : "+obj.noOfPaths(obj.arr,0,0));
	}

	int noOfPaths(int a[][], int i,int j)
	{	
		if(i==m-1 && j==n-1)
			return a[i][j];
		else if (i==m-1)
			return a[i][j+1];
		else if (j==n-1)
			return a[i+1][j];
		else if(a[i][j]==1)
			return noOfPaths(a,i+1,j) + noOfPaths(a,i,j+1);
		else
			return 0;
	}
}