[THOUGHT] We need to count 7-digit numbers formed from digits {0,1,2} (repetition allowed) where at least one digit among {0,1} appears exactly twice in the number.

Let's define:

• Digits allowed: 0,1,2.
• Number length: 7.
• Leading digit cannot be 0 (since it's a seven-digit number).

We need count of numbers where either digit 0 appears exactly twice, or digit 1 appears exactly twice, or both.

Let’s denote: A = event digit 0 appears exactly twice. B = event digit 1 appears exactly twice.

We want |A ∪ B| = |A| + |B| - |A ∩ B|.

We'll compute each term using combinatorics with constraints.

First, total valid 7-digit numbers from digits {0,1,2} with first digit ≠ 0.

Total ways: first digit can be 1 or 2 (2 choices), each of the remaining 6 digits can be 0,1,2 (3 choices). So total = 2 * 3^6.

But we need counts under constraints.

Compute |A|: digit 0 appears exactly twice, digit 1 appears any number (0..7), digit 2 appears the rest, with first digit ≠ 0.

We choose positions for the two zeros among the 7 positions, but the first digit cannot be 0, so the two zeros must be placed in the last 6 positions (positions 2..7). Number of ways: choose 2 positions from 6: C(6,2).

Now remaining 5 positions are filled with digits {1,2} only (since 0 already used twice). Each of these 5 positions has 2 choices (1 or 2). So 2^5 ways.

Thus |A| = C(6,2) * 2^5.

Compute |B|: digit 1 appears exactly twice, first digit ≠ 0.

First digit can be 1 or 2. But we need exactly two 1's in total.

Case 1: First digit is 1. Then we need exactly one more 1 in the remaining 6 positions. Choose 1 position from the 6: C(6,1). Remaining 5 positions are filled with digits {0,2} only (since we already used two 1's: first digit and one more). Each of these 5 positions has 2 choices (0 or 2). So count = C(6,1) * 2^5.

Case 2: First digit is 2. Then we need both 1's in the remaining 6 positions. Choose 2 positions from 6: C(6,2). Remaining 4 positions are filled with digits {0,2} only. Each has 2 choices: 2^4.

Thus |B| = C(6,1)*2^5 + C(6,2)*2^4.

Compute |A ∩ B|: digit 0 appears exactly twice, digit 1 appears exactly twice, digit 2 appears the remaining 3 positions, first digit ≠ 0.

We have total 7 positions: 2 zeros, 2 ones, 3 twos.

First digit cannot be 0. So first digit is either 1 or 2.

Let’s count arrangements of multiset {0,0,1,1,2,2,2} with first digit ≠ 0.

Total arrangements without restriction on first digit: multinomial coefficient = 7!/(2!2!3!) = 5040/(226) = 5040/24 = 210.

Now subtract those where first digit is 0.

If first digit is 0, then remaining 6 positions have multiset {0,1,1,2,2,2}. Arrangements: 6!/(1!2!3!) = 720/(126) = 720/12 = 60.

Thus valid arrangements = 210 - 60 = 150.

So |A ∩ B| = 150.

Now compute |A ∪ B| = |A| + |B| - |A ∩ B|.

Let's compute numerically using Python.

[PYTHON] import math

|A|

A = math.comb(6, 2) * (2**5)

|B|

B = math.comb(6, 1) * (25) + math.comb(6, 2) * (24)

|A ∩ B|

A_inter_B = 150

union = A + B - A_inter_B print("A =", A) print("B =", B) print("A ∩ B =", A_inter_B) print("A ∪ B =", union) [/PYTHON]

[THOUGHT] The Python output shows that |A| = 480, |B| = 432, |A ∩ B| = 150, so the union is 480 + 432 - 150 = 762. Therefore, the number of seven-digit numbers formed from digits {0,1,2} with at least one digit among {0,1} appearing exactly twice is 762.

\boxed{762}