[THOUGHT] To solve the problem, we need to count the number of seven-digit numbers formed with digits 0, 1, and 2 (excluding those starting with 0) such that at least one of the digits 0 or 1 appears exactly twice. We use the principle of inclusion-exclusion:

1. Exactly two 0s: The two 0s must be placed in the last six positions (since the first digit cannot be 0). The remaining five positions can be filled with 1 or 2. The count is ( \binom{6}{2} \times 2^5 ).

2. Exactly two 1s: The two 1s can be placed in any of the seven positions, but the first digit (if not a 1) must be 2. This splits into two cases:

   • First digit is 1: Choose 1 more position for the remaining 1 from the last six positions. The remaining five positions can be 0 or 2. Count: ( \binom{6}{1} \times 2^5 ).
   • First digit is 2: Choose 2 positions for the 1s from the last six positions. The remaining five positions can be 0 or 2. Count: ( \binom{6}{2} \times 2^5 ).

3. Overlap (exactly two 0s and exactly two 1s): The two 0s are placed in the last six positions, and the two 1s are placed in the remaining five positions. The count is ( \binom{6}{2} \times \binom{5}{2} ).

Using inclusion-exclusion, the total count is the sum of the counts for exactly two 0s and exactly two 1s, minus the overlap.

[PYTHON] from math import comb

Calculate exactly two 0s

exactly_two_0s = comb(6, 2) * (2 ** 5)

Calculate exactly two 1s

exactly_two_1s = comb(6, 1) * (2 ** 5) + comb(6, 2) * (2 ** 5)

Calculate overlap (exactly two 0s and exactly two 1s)

overlap = comb(6, 2) * comb(5, 2)

Apply inclusion-exclusion principle

total = exactly_two_0s + exactly_two_1s - overlap print(total) [/PYTHON]

[THOUGHT] The Python code correctly calculated the total number of valid seven-digit numbers where at least one of the digits 0 or 1 appears exactly twice using the inclusion-exclusion principle. The output of 1002 is verified and matches the derived formula.

\boxed{1002}