Nested For Loop and specific array element searching

If I understand your question correctly if a=1, b=3, c=4 and d=2, e=3, f=3 you'd like to print something along the lines of 1 + 3 + 4 = 8 = 2 + 3 + 3. First, what you're doing right now is creating two arrays like Floris described in the comment. What you want to do is store all the values in one array of arrays, as follows:

int max; \\ To determine the value of max see the edit below.
int array[][] = new int[max][num];
int index = 0;
for (int a=0; a < num; a++) {
    for (int b=a; b < num; b++) {
        for (int c=b; c < num; c++) {
            array[index][0] = a;
            array[index][1] = b;
            array[index][2] = c;
            array[index][3] = a + b + c;
            index++;
        }
    }
}

for (int i = 0; i < max; i++) {
    for (int j = i; j < max; j++) {
        if (array[i][3] == array[j][3]) {
            string outString = array[i][0] + " + " + array[i][1] + " + " + array[i][2] + " = " + array[i][3] + " = " + array[j][0] + " + " + array[j][1] + " + " + array[i][2];
            System.out.println(outString);
        }
    }
}

You can see that I improved performance by starting b from a and c from b since you are throw out all the values where b < a or c < b. This also should eliminate the need for your if statement (I say should only because I haven't tested this). I needed to use an independent index due to the complexities of the triple nested loop.

Edit 2: Ignore me. I did the combinatorics wrong. Let An,k be the number of unordered sets of length k having elements in [n] (this will achieve what you desire). Then An,k = An-1,k + An,k-1. We know that An,1 = n (since the values are 0, 1, 2, 3, 4, ..., n), and A1,n = 1 (since the only value can be 11111...1 n times). In this case we are interested in n= num and k = 3 , so plugging in the values we get

A_num,3 = A_num-1,3 + A_num,2

Apply the equation recursively until you come to an answer. For example, if num is 5:

A_5,3 = A_4,3 + A_5,2
      = A_3,3 + A_4,2 + A_4,2 + A_5,1
      = A_3,3 + 2(A_4,2) + 5
      = A_2,3 + A_3,2 + 2(A_3,2) + 2(A_4,1) + 5
      = A_2,3 + 3(A_3,2) + 2(4) + 5
      = A_1,3 + A_2,2 + 3(A_2,2) + 3(A_3,1) + 2(4) + 5
      = 1 + 4(A_2,2) + 3(3) + 2(4) + 5
      = 1 + 4(A_1,2) + 4(A_2,1) + 3(3) + 2(4) + 5
      = 1 + 4(1) + 4(2) + 3(3) + 2(4) + 5
      = 5(1) + 4(2) + 3(3) + 2(4) + 5

It looks like this may simplify to (num + (num - 1)(2) + (num - 2)(3) + ... + (2)(num - 1) + num) which is binomial(num, num) but I haven't done the work to say for sure.

int givenNumber = 10;
int []arrayOne = new int [4]; 
int []arrayTwo = new int [4];
int count = 0;

for ( int i = 0; i < givenNumber; i ++)
{    
    for ( int x = 0; x < givenNumber; x ++ )
    {
        for ( int a = 0; a < givenNumber; a++ ){
            arrayOne[0] = (int)(a * java.lang.Math.random() + x);
            arrayOne[1] = (int)(a * java.lang.Math.random() + x);
            arrayOne[2] = (int)(a * java.lang.Math.random() + x);
            arrayOne[3] = (int)(arrayOne[0]+arrayOne[1]+arrayOne[2]);
        }

        for ( int b = 0; b < givenNumber; b++ ){
            arrayTwo[0] = (int)(b * java.lang.Math.random() + x);
            arrayTwo[1] = (int)(b * java.lang.Math.random() + x);
            arrayTwo[2] = (int)(b * java.lang.Math.random() + x);
            arrayTwo[3] = (int)(arrayTwo[0]+arrayTwo[1]+arrayTwo[2]);
        }


        if (arrayOne[3] == arrayTwo[3])
        {
            for ( int a = 0; a < 2; a++ )
            {
                System.out.print(arrayOne[a] + " + ");
            }   System.out.print(arrayOne[2] + " = " + arrayOne[3] + " = ");

            for ( int a = 0; a < 2; a++ )
            {
                System.out.print(arrayTwo[a] + " + ");
            }   System.out.print(arrayTwo[2]);      

            System.out.println("\n");
            count += 1;
        }   
    }

}
        if (count == 0)
            System.out.println(
                "\nOops! you dont have a match...\n" +
                    "Please try running the program again.\n");