I need some helping starting this discrete math using java. I'm at a complete loss

Question

Your task is to generate all possible functions from X={a, b, c} to a set Y. Set Y contains integers 1, …, n for some integer n≥1. The value of n is provided by the user. To define a function you need to specify what it outputs for each element of X. For instance, if Y= {1, 2}, then f(a)=1, f(b)=2, f(c)=2 defines function f from X to Y

The value of n is provided by the user. To define a function you need to specify what it outputs for each element of X. For instance, if Y= {1, 2}, then f(a)=1, f(b)=2, f(c)=2 defines function f from X to Y.

Write a program that prompts the user to enter the size of Y, then generates, enumerates, and prints out in a neat format all possible functions from X to Y. Your program should number generated functions f1, f2, f3, f4, etc. For each generated function, output whether or not it is one-to-one, onto, or a bijection. Compute total number of functions generated, how many of them are one-to-one, how many of them are onto, and how many of them are bijections.

The program will generate all functions from X={a,b,c} to Y={1,…,n}.

Please enter the value of n: 2

f1(a)=1 f1(b)=1 f1(c)=1

f1 is not one-to-one, not onto, and not a bijection

f2(a)=1 f2(b)=1 f2(c)=2

f2 is not one-to-one, onto, and not a bijection.

f3(a)=1 f3(b)=2 f3(c)=1

f3 is not one-to-one, onto, and not a bijection.

f4(a)=1 f4(b)=2 f4(c)=2

f4 is not one-to-one, onto, and not a bijection.

f5(a)=2 f5(b)=1 f5(c)=1

f5 is not one-to-one, onto, and not a bijection.

f6(a)=2 f6(b)=1 f6(c)=2

f6 is not one-to-one, onto, and not a bijection.

f7(a)=2 f7(b)=2 f7(c)=1

f7 is not one-to-one, onto, and not a bijection.

f8(a)=2 f8(b)=2 f8(c)=2

f8 is not one-to-one, not onto, and not a bijection.

There are 8 functions total.

0 of them are one-to-one.

6 of them are onto.

0 of them are bijections.

Answer 1

public class Main {

    public static void main(String[] args) {
        Set<Integer> yGroup = new HashSet<>();
        int n;

        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter value of n:");
        n = Integer.parseInt(scanner.nextLine());
        scanner.close();

        for (int i = 1; i <= n; i++) {
            yGroup.add(i);
        }

        int nf = 1;
        int oneToneCounter = 0;
        int ontoCounter = 0;
        int bijectCounter = 0;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                for (int k = 1; k <= n; k++) {
                    System.out.print("f" + nf + "(a)=" + i + " ");
                    System.out.print("f" + nf + "(b)=" + j + " ");
                    System.out.print("f" + nf + "(c)=" + k + " ");
                    System.out.println();

                    boolean injective = false;
                    if ((i == j) || (i == k) || (j == k)) {
                        System.out.print("f" + nf + " is not a one to one. ");
                    } else {
                        System.out.print("f" + nf + " is a one to one. ");
                        oneToneCounter++;
                        injective = true;
                    }

                    yGroup.remove((Integer) i);
                    yGroup.remove((Integer) j);
                    yGroup.remove((Integer) k);

                    boolean onto = false;
                    if (yGroup.isEmpty()) {
                        System.out.print("Is an onto. ");
                        onto = true;
                        ontoCounter++;
                    } else {
                        System.out.print("Is not an onto. ");
                    }

                    yGroup.add((Integer) i);
                    yGroup.add((Integer) j);
                    yGroup.add((Integer) k);

                    if (injective && onto) {
                        System.out.print("Is bijective. ");
                        bijectCounter++;
                    } else {
                        System.out.print("Is not bijective. ");
                    }

                    System.out.println();
                    System.out.println();
                    nf++;
                }
            }
        }

        System.out.println();
        System.out.println("There are " + nf + " functions total");
        System.out.println(oneToneCounter + " of them are one-to-one");
        System.out.println(ontoCounter + " of them are onto");
        System.out.println(bijectCounter + " of them are bijections");
    }

}