Question: So i'm trying to generate all binaries of a size n but with the condition that only k 1s. i.e

for size n = 4, k=2, (there is 2 over 4 combinations)

```
1100
1010
1001
0110
0101
0011

```

I'm stuck and can't figure out how to generate this.

Answer: 
Here's my non-recursive take on this algorithm. Because there are `2^n` permutations of binary strings, we can use a for-loop to iterate through every possible string and check if the amount of "1"s is not equal to `k`:

```
private static void generate(int n, int k) {
    for (int i = 0; i < Math.pow(2, n); i++) {
        if (Integer.bitCount(i) != k) {
            continue;
        }

        String binary = Integer.toBinaryString(i);

        if (binary.length() < n) {
            System.out.format("%0" + (n - binary.length()) + "d%s\n", 0, binary);
        } else {
            System.out.println(binary);
        }
    }
}

```
int n = 4, k=2;

```
    for (int i = 0; i < Math.pow(2,n) ; i++) {
        int a = Integer.bitCount(i);
        if (a == k) System.out.println(Integer.toBinaryString(i));
    }

```

I think this is the simplest answer.