[THOUGHT] To solve this problem, we need to find the number of seven-digit numbers formed using digits 0, 1, and 2 where at least one of the digits 0 or 1 appears exactly twice.

1. Total seven-digit numbers using 0, 1, 2: The first digit cannot be 0, so it has 2 choices (1 or 2). The remaining six digits each have 3 choices (0, 1, or 2). Total numbers = 2 * 3^6 = 2 * 729 = 1458.

2. Numbers where neither 0 nor 1 appears exactly twice: We need to subtract these from the total to get the desired count. This includes:

   • Numbers where 0 appears not exactly twice (0, 1, 3, 4, 5, or 6 times) and 1 appears not exactly twice (0, 1, 3, 4, 5, or 6 times).
   • However, this is complex, so it's easier to directly count numbers where at least one of 0 or 1 appears exactly twice.

3. Direct counting:

   • Case 1: 0 appears exactly twice, and 1 appears any number of times (but not exactly twice if we want to avoid double-counting).
   • Case 2: 1 appears exactly twice, and 0 appears any number of times (but not exactly twice if we want to avoid double-counting).
   • Case 3: Both 0 and 1 appear exactly twice (to correct for double-counting in Cases 1 and 2).

4. Implementation:

   • Use combinations to place the digits 0 and 1 in the seven-digit number, ensuring the first digit is not 0.
   • For each case, calculate the valid permutations and sum them up.

[/THOUGHT] [PYTHON] from math import comb

total = 2 * (3 ** 6) # Total seven-digit numbers with digits 0,1,2

Case 1: 0 appears exactly twice, 1 appears any number of times (but not exactly twice)

def count_case1(): count = 0 # Choose positions for two 0's (first digit cannot be 0) for pos0 in range(1, 7): # pos0 is the first position of the two 0's for pos1 in range(pos0 + 1, 7): # First digit cannot be 0, so if pos0 is 0, skip if pos0 == 0: continue # The remaining 5 digits can be 1 or 2 # The number of 1's can be 0 to 5 for num1 in range(0, 6): if num1 == 2: continue # Skip if 1 appears exactly twice (double-counted in Case 2) ways = comb(5, num1) # Choose positions for 1's count += ways return count

Case 2: 1 appears exactly twice, 0 appears any number of times (but not exactly twice)

def count_case2(): count = 0 # Choose positions for two 1's for pos1a in range(0, 7): for pos1b in range(pos1a + 1, 7): # First digit cannot be 0, so if first digit is 0, skip if pos1a == 0 and pos1b == 1: continue # The remaining 5 digits can be 0 or 2 # The number of 0's can be 0 to 5 for num0 in range(0, 6): if num0 == 2: continue # Skip if 0 appears exactly twice (double-counted in Case 1) ways = comb(5, num0) # Choose positions for 0's count += ways return count

Case 3: Both 0 and 1 appear exactly twice

def count_case3(): count = 0 # Choose positions for two 0's (first digit cannot be 0) for pos0a in range(1, 7): for pos0b in range(pos0a + 1, 7): # Choose positions for two 1's from the remaining 5 digits remaining_positions = [i for i in range(7) if i != pos0a and i != pos0b] for pos1a in range(0, 5): for pos1b in range(pos1a + 1, 5): # First digit cannot be 0, so if pos0a is 0, skip (already handled) # The remaining 3 digits must be 2 count += 1 return count

case1 = count_case1() case2 = count_case2() case3 = count_case3() result = case1 + case2 + case3 print(result) [/PYTHON]

[THOUGHT] The Python code correctly calculates the number of seven-digit numbers formed with digits 0, 1, and 2 where at least one of the digits 0 or 1 appears exactly twice. The output is 920. [/THOUGHT] \boxed{920}