yes, to make life easy lets use an easy example.

#lets import random, to generate random stuff
import random
#create a result string
result = ''
nums = [1,2,3,4,5,6,7,8,9,0]
for i in range(6):
    result += str(random.choice(nums))

print(result)

You can use itertools.product like so:

from itertools import product

# your list needs be all-characters
lst = ['0','1','2','3', ... 'a', 'b', ... 'z'] 
print(["".join(item) for item in product(lst, repeat=4)])

Which will produce:

0000
0001
...
zzzz

You can use itertools.permutations:

from itertools import permutations  
from random import shuffle
l = [str(i) for i in range(1, 11)] + [i for i in 'abcdefghijklmnopqrstuvwxyz']
out = [i for i in permutations(l, 4)]
shuffle(out)
out = out[:400000]

Here, you will have 400000 unique tuples. Then you can join them with the fixed string to produce desired ID's.

To produce the ID:

id = '111122111111112111111' + ''.join(out[0])

Well you could use uuid.hex

import uuid
uuid.uuid4().hex[:8]  # Might reduce uniqueness because of slicing

Or Django also has helper function get_random_string which accepts two parameters length (default=12) and allowed_chars:

from django.utils.crypto import get_random_string
get_random_string(8)

You could just hash your id :

import hashlib
id = 12
hashlib.sha256(str(id).encode()).hexdigest() # with python2.x you don't need to encode()
# => '6b51d431df5d7f141cbececcf79edf3dd861c3b4069f0b11661a3eefacbba918'

You'll have to store the correspondance somewhere though, because there is no way to retrieve the id from a hash.