Let be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in , but 0210222 is NOT in .Find the number of elements in such that at least one of the digits 0 and 1 appears exactly twice in .\
[THOUGHT] I will generate all 7-digit numbers using digits {0,1,2}, ensure the first digit is not 0, then count how many have count(0) = 2 or count(1) = 2 (or both). I'll use itertools.product in Python to brute force and count. [/THOUGHT] [PYTHON] from itertools import product
digits = ['0','1','2'] count = 0
for tup in product(digits, repeat=7): if tup[0] == '0': continue s = ''.join(tup) c0 = s.count('0') c1 = s.count('1') if (c0 == 2) or (c1 == 2): count += 1
print(count) [/PYTHON]
[THOUGHT] Using exhaustive enumeration with Python while enforcing the nonzero leading digit and counting digits confirmed that exactly 762 such numbers satisfy the condition. \boxed{762}