You only have to use a single for loop - like this:

public static void main(String[] args) 
{
    int[] numbers = new int[10];       
    //Generates 10 Random Numbers in the range 1 -20
    for(int i = 0; i < numbers.length; i++) {
      numbers[i] = (int)(Math.random()*20 + 1);
    }//end for loop
    System.out.println("Numbers Generated: " + Arrays.toString(numbers));
}

public class RandomNumbers {

  public static void main(String[] args) {
    int[] randomNumbers = new int[] { 2, 3, 5, 7, 11, 13, 17, 19 };

    Random r = new Random();
    int nextRandomNumberIndex = r.nextInt(randomNumbers.length);
    System.out.println(randomNumbers[nextRandomNumberIndex]);
  }

}

For those still looking for a solution:

In Java 8 or later, this can be answered trivially using Streams without any loops or additional libraries.

int[] range = IntStream.rangeClosed(1, 10).toArray();

This will produce an array with the integers from 1 to 10.

A more general solution that produces the same result is below. This can be made to produce any sequence by modifying the unary operator.

int[] range = IntStream.iterate(1, n -> n + 1).limit(10).toArray();

First, let's create the set:

Set<String> set = new HashSet<>();

Now, let's generate 5 random strings. There are many ways to do this, here is an answer about it. But since you did not ask for a specific length, let's proceed with this:

String randomStr = Long.toHexString(Double.doubleToLongBits(Math.random()));

Now, let's repeat the random generation 5 times:

for (int i = 0; i < 5; i++) {
    set.add(Long.toHexString(Double.doubleToLongBits(Math.random())));
}

Now, the problem is that this does not guarantee the set will have 5 random strings, as we may have similar ones. To ensure that, we should do:

while (set.size() < 5) {
    set.add(Long.toHexString(Double.doubleToLongBits(Math.random())));
}

The code above will continue generating random strings until the set contains at least 5. I'm not checking if it already contains the string because a Set, by definition, contains no duplicate elements. So, adding a duplicate won't increase the size.

1) fill an array with sequential values

2) shuffle the array