Задача с выводом на печать 252 строчек

for x in range(252):
    import random
    from random import shuffle
    from random import sample
    list = [1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
    a = random.sample(list, 10)
    print(*a, sep = '')

from random import shuffle
 
l = [1] * 5 + [0] * 5
p = []
for x in range(252):
    shuffle(l)
    while l in p:
        shuffle(l)
    p.append(l[:])
    print(*l, sep='')

### 1
from random import shuffle
xset = set()
lst = [1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
while len(xset) < 252:
    shuffle(lst)
    xset.add(tuple(lst))
for x in xset:
    print(*x)
print()
 
### 2
i = k = 0
while k < 252:
    s = bin(i)[2:].rjust(10, '0')
    if s.count('1') == 5:
        print(' '.join(s))
        k += 1
    i += 1

for x in range(252):
    import random
    from random import shuffle
    from random import sample
    list = [1, 0, 1, 0, 1, 0, 1, 0, 1, 0]
    a = random.sample(list, 10)
    print(*a, sep = '')

ss = set()
i = k = 0
while i < 1024:
    s = bin(i)[2:].rjust(10, '0')
    if s.count('1') == 5 :
        ss.add(''.join(s))
        k += 1
    i += 1
print(k)

print(*("".join(i) for i in __import__("itertools").permutations("0000011111", 10)), sep="\n")

print(*("".join(i) for i in sorted(set(__import__("itertools").permutations("0000011111", 10)))[:252]), sep="\n")

print(*("".join(i) for i in sorted(set(__import__("itertools").permutations("0000011111", 10)))), sep="\n")

print(*("".join(i) for i in set(__import__("itertools").permutations("0000011111", 10))), sep="\n")

def next_permutation(a):
    """Generate the lexicographically next permutation inplace.
 
    https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
    Return false if there is no next permutation.
    """
    # Find the largest index i such that a[i] < a[i + 1]. If no such
    # index exists, the permutation is the last permutation
    for i in reversed(range(len(a) - 1)):
        if a[i] < a[i + 1]:
            break  # found
    else:  # no break: not found
        return False  # no next permutation
 
    # Find the largest index j greater than i such that a[i] < a[j]
    j = next(j for j in reversed(range(i + 1, len(a))) if a[i] < a[j])
 
    # Swap the value of a[i] with that of a[j]
    a[i], a[j] = a[j], a[i]
 
    # Reverse sequence from a[i + 1] up to and including the final element a[n]
    a[i + 1:] = reversed(a[i + 1:])
    return True
a = [0] * 5 + [1] * 5
for _ in range(252):
    next_permutation(a)
    print(*a, sep="")