How to get all possible combinations from two arrays?

While @TimBiegeleisen solution would work like a charm, its complexity might be an issue. The better approach would be a code like this:

static void combinationUtil(
        int[] arr, int n, int r, int index, int[] data, int i) {
    // Current combination is ready to be printed, print it 
    if (index == r) {
        for (int j = 0; j < r; j++)
            System.out.print(data[j] + " ");
        System.out.println("");
        return;
    }

    // When no more elements are there to put in data[] 
    if (i >= n)
        return;

    // current is included, put next at next location 
    data[index] = arr[i];
    combinationUtil(arr, n, r, index + 1, data, i + 1);

    // current is excluded, replace it with next (Note that 
    // i+1 is passed, but index is not changed) 
    combinationUtil(arr, n, r, index, data, i + 1);
}

// The main function that prints all combinations of size r 
// in arr[] of size n. This function mainly uses combinationUtil() 
static void printCombination(int arr[], int n, int r) {
    // A temporary array to store all combination one by one 
    int data[] = new int[r];

    // Print all combination using temprary array 'data[]' 
    combinationUtil(arr, n, r, 0, data, 0);
}

Source: GeeksForGeeks and my IDE :)

Use a triple loop:

for (int i=0; i < operators.length; ++i) {
    for (int j=0; j < operators.length; ++j) {
        for (int k=0; k < operators.length; ++k) {
            System.out.println(numbers[0] + operators[i] + numbers[1] + operators[j] +
                numbers[2] + operators[k] + numbers[3]);
        }
    }
}

You essentially want to take the cross product of the operators vector (if it were a vector). In Java, this translates to a triply-nested set of loops.