How do I find all possible positions?

Question

If I have 3 positions and 3 numbers I know there are 6 possible soultions (of having a diff number in each)
If I have 4 positions and 4 numbers I know there are 12 possible soultions (of having a diff number in each)
I know the function to find the number of diff positions in n! where n is the number of positions ( + is the ! called shreek?)
5! = 5 * 4 * 3 * 2 * 1 = 120 diff positions
But how do I find out what all the different positions are (ie to print out a list)
I just can't work out how to write an algorithm for this.
Maybe there is somwwhere on the web I can find out, but Google won't let me search for ! + I don't know the real name for it.

Answer 1

I dashed this off. I can't read the one posted above without my eyes bleeding. This one uses recursion.
import java.util.*;
public class Test
    public static void main(String[] args)
        for (int[] array : permutations(5)) print(array);
    static void print(int[] array)
        for (int i = 0; i < array.length; i++) {
            System.out.print(array);
System.out.print(' ');
System.out.println();
static List<int[]> permutations(int number)
int[] input = new int[number];
for (int i = 1; i <= number; i++) input[i - 1] = i;
return permutations(input, 0);
static List<int[]> permutations(int[] input, int from)
List<int[]> bigPerm = new ArrayList<int[]>();
for (int i = from; i < input.length; i++) {
int[] next = new int[input.length];
System.arraycopy(input, 0, next, 0, from);
next[from] = input[i];
for (int j = from + 1; j < input.length; j++) {
next[j] = input[(j <= i) ? j - 1 : j];
if (from < input.length - 1) {
bigPerm.addAll(permutations(next, from + 1));
} else {
bigPerm.add(next);
return bigPerm;