Generating permutations in Java using array

The following code generates all possible permutations till the entered integer integers. Similar program can be made to permute the characters of a string.

Explanation:
First we enter only 1 in a array.
Then for entering 2 we have two choices- before 1 and after 1 as- "21" and "12".
The program uses recursion so for "21" we can add 3 at 3 different places as "321", "231" and "213" i.e., at first position, between them and in the last position.

The code is given below: 

import java.util.Arrays;
import java.util.Scanner;

/**
 *
 * @author Aditya Vijayvergia @ thecodingedge.blogspot.com
 */
public class GeneratingPermutations {

    static int perm(int []a,int n,int m)
    {                          //n=the element that is being added in this recursive call
        int []b=new int[m];    //m=total no of elemets
        for(int i=0;i<n;i++)
            {
                System.arraycopy(a, 0, b, 0, m);//copy the state of array a to array b
                for(int j=n-1;j>i;j--)//shift values 1 step ahead
                    b[j]=b[j-1];
                b[i]=n;//store n at the empty location
                if(n==m)
                {
                    System.out.println(Arrays.toString(b));
                }
                else{    
                    perm(b,n+1,m);   
                    }
            }
        return 0;
    }
    
    public static void main(String[] args) {
        int m;
        System.out.print("enter the number of elements:");
        Scanner in=new Scanner(System.in);
        m=in.nextInt();
        int []a=new int[m];
        perm(a,1,m);
    }
}

Sample Output: