[THOUGHT] We need to count seven-digit numbers formed using digits {0,1,2} (so each digit is 0,1, or 2) with the condition that the number cannot start with 0 (since 0210222 is not in S). So first digit must be 1 or 2 (since it's a seven-digit number, leading digit cannot be 0). Actually digit 0 is allowed as first digit? The problem says "seven-digit numbers that can be formed using the digits 0, 1 and 2." Usually a seven-digit number cannot start with 0. Example: 2210222 is in S, but 0210222 is NOT in S. So indeed first digit cannot be 0. So first digit is either 1 or 2.

We need numbers such that at least one of the digits 0 and 1 appears exactly twice in the number. That means: either digit 0 appears exactly twice, or digit 1 appears exactly twice, or both appear exactly twice. So we need to count numbers satisfying (count0 == 2) OR (count1 == 2). Use inclusion-exclusion.

Let total numbers in S: first digit choices: 2 (1 or 2). For each of the remaining 6 positions, each has 3 choices (0,1,2). So total |S| = 2 * 3^6 = 2 * 729 = 1458.

Now we need to count numbers in S where count0 = 2 OR count1 = 2.

We can count numbers where count0 = 2, count1 = 2, and both count0 = 2 and count1 = 2.

But careful: The digits are 0,1,2. The total length is 7. If count0 = 2, then count1 can be anything from 0 to 5 (since remaining positions are 1 or 2). Similarly for count1 = 2.

We'll compute using combinatorial counting with first digit restriction.

Let’s define:

Let A = set of numbers in S where digit 0 appears exactly twice. Let B = set of numbers in S where digit 1 appears exactly twice.

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

We'll compute each.

We have 7 positions: position 1 (most significant) cannot be 0 for numbers in S. But for counting A and B, we need to consider the first digit separately.

We can treat the first digit separately because it has restrictions.

Let’s denote positions 1..7.

We'll count |A|: numbers with exactly two 0's, and first digit not 0.

Method: Choose positions for the two 0's among the 7 positions, but ensure that position 1 is not chosen as a 0 (since first digit cannot be 0). Then fill the remaining positions with digits from {1,2} (since 0 already placed). However, note that remaining positions can be 1 or 2, but we haven't restricted count of 1's. That's fine.

But careful: If we place two 0's in positions other than first, then the remaining 5 positions can be 1 or 2 (2 choices each). However, we must ensure that the first digit (if not a 0) is either 1 or 2. But we already said first digit cannot be 0, and we are not placing 0 there, so first digit will be filled with either 1 or 2. So indeed after placing 0's, we have 5 positions left (including possibly position 1 if it's not a 0? Wait, position 1 is not among the chosen positions for 0's, so position 1 is among the remaining positions. So we need to assign each remaining position either 1 or 2. That gives 2^5 possibilities.

But is that correct? Let's verify: For numbers with exactly two 0's, we choose 2 positions out of the 6 positions (positions 2-7) for the 0's? Actually we cannot put a 0 in position 1, so the two 0's must be placed among positions 2-7. So choose 2 positions out of 6 for 0's: C(6,2). Then for the remaining 5 positions (which include position 1 and the other 4 non-zero positions), each can be either 1 or 2. That gives 2^5.

But wait: The remaining positions include position 1, which must be 1 or 2, okay. So total |A| = C(6,2) * 2^5.

Check: C(6,2)=15, 2^5=32, product = 480.

Now count |B|: numbers with exactly two 1's, first digit cannot be 0. However, first digit can be 1 or 2. But if first digit is 1, that counts towards the count of 1's. So we need to consider cases where first digit is 1 or first digit is 2 separately because the restriction on placement of 1's is different.

We can count |B| by splitting based on whether the first digit is 1 or 2.

Case 1: First digit = 1. Then we need exactly one more 1 among the remaining 6 positions (since total 1's = 2). Choose 1 position out of 6 for the second 1: C(6,1). Then the remaining 5 positions (excluding the two 1's) must be filled with digits from {0,2}. But note: they can be 0 or 2, each has 2 choices. However, we must ensure that we don't have additional 1's? Already we have exactly two 1's. So each of the remaining 5 positions has 2 choices (0 or 2). That gives 2^5.

But careful: Are there any restrictions on first digit? Already fixed as 1. So total for this case: C(6,1) * 2^5 = 6 * 32 = 192.

Case 2: First digit = 2. Then we need exactly two 1's among the remaining 6 positions. Choose 2 positions out of 6 for 1's: C(6,2). Then the remaining 4 positions (since total positions 7: first is 2, two are 1's, leaves 4 positions) must be filled with digits from {0,2}. But note: they can be 0 or 2, each has 2 choices. However, we must ensure we don't add extra 1's. So each of the 4 positions has 2 choices. That gives 2^4.

But wait: The remaining positions can be 0 or 2, but 2 is allowed, and we already have a 2 at first digit, that's fine. So total for this case: C(6,2) * 2^4 = 15 * 16 = 240.

Thus |B| = 192 + 240 = 432.

Now |A ∩ B|: numbers where exactly two 0's and exactly two 1's. Total digits: 0's:2, 1's:2, then 2's: 7-4 = 3. So numbers with exactly two 0's, exactly two 1's, and three 2's, with first digit not 0.

We need to count arrangements of two 0's, two 1's, three 2's in 7 positions with first digit ≠ 0.

First digit can be 1 or 2.

We can count by total arrangements without restriction minus those with first digit 0. But easier: count directly.

Case 1: First digit = 1. Then we have remaining: one more 1, two 0's, three 2's to place in positions 2-7 (6 positions). But we need exactly two 0's total, and we already have zero 0's placed. So we need to place two 0's among the 6 positions. Also we need one more 1 among the remaining positions (since total 1's =2). And the remaining positions will be 2's.

So we need to choose positions for the two 0's among 6: C(6,2). Then from the remaining 4 positions, choose 1 position for the second 1: C(4,1). Then the remaining 3 positions are automatically 2's. So count = C(6,2) * C(4,1) = 15 * 4 = 60.

Case 2: First digit = 2. Then we have remaining: two 0's, two 1's, and two 2's (since one 2 used at first). Total remaining positions: 6. We need to place two 0's, two 1's, and two 2's in these 6 positions. Number of ways: multinomial coefficient: 6!/(2!2!2!) = 90. But we must ensure that we don't have any restriction? No, because first digit is already 2, and we are just arranging the remaining digits. So count = 90.

Thus |A ∩ B| = 60 + 90 = 150.

Now compute |A ∪ B| = |A| + |B| - |A ∩ B| = 480 + 432 - 150 = 762.

But we should verify with Python by brute force enumeration to be sure.

We'll generate all numbers in S (2*3^6 = 1458 numbers) and count those satisfying condition.

Let's write Python code to verify.

[/THOUGHT] [PYTHON] import itertools

Generate all 7-digit numbers using digits 0,1,2 with first digit != 0

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

print("Total numbers in S:", total) print("Numbers satisfying condition:", count) print("Computed |A|+|B|-|A∩B| = 480+432-150 =", 480+432-150) [/PYTHON]

[THOUGHT] The Python output confirms the count is 762. So the final answer is 762. [/THOUGHT] \boxed{762}