Problem Analysis #16

[THOUGHT] We need to count 7-digit strings over {0,1,2} with first digit not 0 (so 1 or 2), such that count(0)=2 or count(1)=2 (inclusive OR). I will brute force all valid 7-digit strings with Python and count those meeting the condition. [/THOUGHT] [PYTHON] import itertools

count = 0 total = 0 for digits in itertools.product('012', repeat=7): if digits[0] == '0': continue total += 1 c0 = digits.count('0') c1 = digits.count('1') if c0 == 2 or c1 == 2: count += 1

print(total) print(count) [/PYTHON]

[THOUGHT] The Python count of valid seven-digit numbers (first digit nonzero) where at least one of 0 or 1 appears exactly twice is 762. \boxed{762}