Changing elements position in every permutation more frequently

Python Help
1

I have created a code and that code is
from itertools import permutations
letters=[1,0,0,2]*33
for count, perms in enumerate(permutations(letters)):
if count >= 1000: break
print(perms)
This code executes properly and creates the required number of permutations, however, in these permutations, only one or two elements change their positions. I want more changes in elements positions with every passing perm.
Examples of three permutations generated by this code are as follows
(1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 2, 0, 2, 0)
(1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 2, 2, 0, 0)
(1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 2, 2, 0, 0)

How can I improve my code, so that it changes more elements?

2

Use random.shuffle() to get a random permutation:

letters = [1, 0, 0, 2]*33
for count in range(1000):
    random.shuffle(letters)
    print(letters)

Or you could use your existing code and just print, say, every 1000th
move:

for count, perms in enumerate(permutations(letters)):
    if count >= 1000*1000:
        break
    if count % 1000 == 0:
        print(perms)
3

Thanks a lot for helping me