bin_perms.py

def next_perm(v):
    """
    Generates next permutation with a given amount of set bits,
    given the previous lexicographical value.
    Taken from http://graphics.stanford.edu/~seander/bithacks.html
    """
    t = (v | ( v - 1)) + 1
    w = t | ((((t & -t) / (v & -v)) >> 1) - 1)
    return w


def perm_gen(size, set_bits):
    start = 2**set_bits - 1
    cur = start
    while True:
        yield cur
        if cur >= start << (size - set_bits):
            break
        cur = next_perm(cur)


print ['{0:010b}'.format(x) for x in perm_gen(10, 3)]