from itertools import product

count = 0

for i in product ('01a', repeat = 8 ):

s = ''.join(i)

if s[0] != '0' and s.count('0') == 2 and s.count('a') <5:

count += 9**s.count('1') * 6**s.count('a')

print(count)

from itertools import product

s = product('123456789ABCDEF',repeat = 6)

k = 0

for i in s:

p = ''.join(i)

if p.count('A')+p.count('B')+p.count('C')+p.count('D')+p.count('E')+p.count('F')<5:

k += 1

print(k*21)# 21 спо­соб по­ста­вить два 0 на 8 мест.