import java.util.*;

public class PermutationSeating2 {

    // We'll use a global to count the number of permutations.
    static int count;

    public static void main (String[] argv)
    {
        // Test case 1: M=3, K=2.
	int numSeats = 3;
	int numPeople = 2;
	int[] seats = new int [numSeats];
	count = 0;
	printPermutations (numSeats, numPeople, seats, 1);
	System.out.println ("  => " + count + " permutations");

        // Test case 1: M=5, K=2.
	numSeats = 5;
	numPeople = 2;
	seats = new int [numSeats];
	count = 0;
	printPermutations (numSeats, numPeople, seats, 1);
	System.out.println ("  => " + count + " permutations");
    }

    static void printPermutations (int numSpaces, int numRemaining, int[] seats, int person)
    {
        // Bottom-out case. Note that we are printing here, since each time 
        // we get here we complete one permutation.
	if (numRemaining == 0) {

	    // Print.
	    System.out.println ( Arrays.toString(seats) );

            // Remember to increment the number of permutations found.
	    count ++;

	    return;
	}


	// Otherwise, non-base case: look for an empty spot for "person"
	for (int i=0; i<seats.length; i++) {
	    if (seats[i] == 0) {

		// Empty spot.
		seats[i] = person;

                // Recursively assign remaining, starting with person+1
		printPermutations (numSpaces-1, numRemaining-1, seats, person+1);

                // Important: we need to un-do the seating for other trials.
		seats[i] = 0;
	    }
	} //end-for
    }

}