Random r = new Random();
int[][] 2dArray = new int[4][4];
for (int i = 0; i<4; i++){
for (int j = 0; j<4; j++){
2dArray[i][j] = r.nextInt(8) + 1;
// nextInt returns a random int >= 0 and < n
}
}

Random r = new Random();
int[][] 2dArray = new int[4][4];
for (int i = 0; i<4; i++){
for (int j = 0; j<4; j++){
2dArray[i][j] = r.nextInt(8) + 1;
// nextInt returns a random int >= 0 and < n
}
}

Thx...
But how I gona to make sure every number from 1 to 8 are come out in a pairs?

Sory, my english not very gud...
I mean a pair of random number is like the program will generate a match of number that from 1 to 8 like the example below...

2 5 1 3
3 8 7 6
1 2 4 5
4 6 8 7

All number must come out in a pair, because I need to write a program that plays the memory matching game.

I don't know if it's the best idea but that's what I got at a late hour :yawn: :

Create an array of size 8 which will hold the times each number 1 .. 8 was generated so far, and initialize it with 0's (we have no generated numbers):

int[] t = new int[8];
for (int i=0; i<8; i++){
t[i] = 0;
}

then when a number is generated increment his corespondent in this array (t[0] for 1, t[1] for 2, ... ). If the number has been generated twice then generate another number, else put it in the 2D array and increment his value in t array.

Random r = new Random();
int[][] 2dArray = new int[4][4];
int number;
for (int i = 0; i<4; i++){
for (int j = 0; j<4; j++){
number = r.nextInt(8) + 1; // generate a number
while (t[number-1] == 2){
// if the number was already generated 2 times
number = r.nextInt(8) + 1; // generate another number
}
// we have a number that was NOT generated 2 times
// add it to 2D array
2dArray[i][j] = number;
// increment the times he was generated
t[number-1] = t[number-1] + 1;
// number-1 because the index of an array starts from 0, not from 1
}
}

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