Java: Generator of true's & false's combinations by giving the number N;

Here is a really basic way using only Java APIs:

final int n = 3;
for (int i = 0; i < Math.pow(2, n); i++) {
    String bin = Integer.toBinaryString(i);
    while (bin.length() < n)
        bin = "0" + bin;
    System.out.println(bin);
}

Result:

000
001
010
011
100
101
110
111

Of course, you can set n to whatever you like. And, with this result, you can pick the nth character from the string as true/false.

If you only need to check if a bit is true, you don't need to convert it to a string. This is just to illustrate the output values.

Just a clue but think about the bits that are set for a number with at most 'n' bits. You'll see if you go from 0 to 'n' number of bits (3 in this case); the bits are 000, 001, 010, 011, 100, 101, 110, 111. You can figure out the max number that can fit in 'n' bits by using the ((n*n)-1) formula.

This should do the trick

int cols = 3;
int rows = (int) Math.pow(2, cols);
for (int row = 0; row < rows; row++)
    System.out.println(String.format("%" + cols + "s", 
            Integer.toBinaryString(row)).replace(' ', '0').replace('1', 'X'));

out:

000
00X
0X0
0XX
X00
X0X
XX0
XXX