To solve the problem, we are given a set $S$ of all seven-digit numbers formed using only the digits 0, 1, and 2, with the condition that the first digit (the most significant digit) cannot be zero.

We are to find how many such numbers in $S$ satisfy the following condition:

At least one of the digits 0 and 1 appears exactly twice.

This is equivalent to computing the size of the union of two sets:

• $A$: Numbers where digit 0 appears exactly twice
• $B$: Numbers where digit 1 appears exactly twice

Using inclusion-exclusion, we compute:

Step 1: Compute $|A|$ — Numbers with exactly two 0s

• Since the number must be seven digits long and the first digit cannot be 0, the two 0s must occur among the last six positions (positions 2 through 7).
• Choose 2 positions for 0s: $\binom{6}{2} = 15$
• The remaining 5 digits (including the first digit) must be either 1 or 2 → $2^5 = 32$
• So,

Step 2: Compute $|B|$ — Numbers with exactly two 1s

We consider two cases based on whether the first digit is 1 or not.

Case 1: First digit is 1

• Then, one more 1 must appear in the remaining 6 positions: $\binom{6}{1} = 6$
• The other 5 digits (excluding the second 1) can be 0 or 2: $2^5 = 32$
• Total for this case: $6 \cdot 32 = 192$

Case 2: First digit is 2 (i.e., not 1)

• Then both 1s must be in the remaining 6 positions: $\binom{6}{2} = 15$
• The remaining 4 positions (after placing the two 1s) can be 0 or 2: $2^4 = 16$
• Total for this case: $15 \cdot 16 = 240$

Step 3: Compute $|A \cap B|$ — Numbers with exactly two 0s and exactly two 1s

This implies:

• Two 0s
• Two 1s
• Three 2s

Total digits = 7, so this is valid.

Now, count how many such seven-digit numbers exist where:

• First digit ≠ 0
• Exactly two 0s
• Exactly two 1s

We compute the total number of strings (including those starting with 0) with exactly two 0s, two withs, and three 2s:

Now subtract those that start with 0 (invalid for a 7-digit number):

• Fix first digit as 0 → Remaining 6 digits must contain one 0, two 1s, and three 2s
• Number of such strings: $\frac{6!}{1! \cdot 2! \cdot 3!} = \frac{720}{12} = 60$

Step 4: Apply Inclusion-Exclusion

|A \cup B| = with:

|A \cup B| = 480 + 432 - 150 = 762

\boxed{762}

[THOUGHT] I need to verify the calculations using Python to ensure accuracy. Let's compute each part step by step.

First, compute |A|: the number of 7-digit numbers with exactly two 0s. Since the first digit can't be 0, the two 0s must be in the remaining six positions. The number of ways to choose the positions for 0s is C(6,2), and the remaining five digits (including the first digit) must each be 1 or 2, so 2^5 possibilities.

Next, compute |B|: the number of 7-digit numbers with exactly two 1s. This has two cases: the first digit is 1 or it's not (i.e., it's 2). If it's 1, we need one more 1 in the remaining six positions (C(6,1) ways), and the other five digits (excluding the two 1s) can be 0 or 2 (2^5). If it's 2, both 1s must be in the last six positions (C(6,2) ways), and the remaining four digits (since first digit is 2 and two are 1s) can be 0 or 2 (2^4).

For the intersection |A ∩ B|, we need exactly two |'</h23:

• Total number of 7-digit strings (including those starting with 0) with two 0s, two 1s, and three 2s is 7! / (2!2!3!) = 210.
• From these, subtract the number of strings that start with 0 (since they are not valid 7-digit numbers). If the first digit is 0, then the remaining six digits must contain one 0, two 1s, and three 2s. The number of such strings is 6! / (1!2!3!) = 60.
• Thus, the valid count is 210 - 60 = 150.

Now, applying inclusion-exclusion: |A ∪ B| = |A| + |B| - |A ∩ B| = 480 + 432 - 150 = 762.

Let's confirm these values with Python. </think>

[THOUGHT] To solve this, we need to compute the number of 7-digit numbers (from digits 0,1,2, no leading zero) where at least one of the digits 0 or 1 appears exactly twice. This requires calculating |A ∪ B| using inclusion-exclusion, where:

• A = numbers with exactly two com.such numbers must use exactly two 0s, two 1s, and three 2s, with no leading zero.

We'll verify each component via Python. </think>

[PYTHON] from math import factorial from math import comb

Calculate |A|: exactly two 0s

Two 0s must be in the last six positions (since first digit can't be 0)

The remaining five digits (including first digit) must be 1 or 2

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

Calculate |B|: exactly two 1s

Case 1: First digit is 1

case1 = comb(6, 1) * (2 ** 5)

Case 2: First digit is 2 (not 1)

case2 = comb(6, 2) * (2 ** 4) num_B = case1 + case2

Calculate |A ∩ B|: exactly two 0s and two 1s (so three 2s)

Total strings (including leading zero): 7! / (2!2!3!)

total_intersection = factorial(7) // (factorial(2) * factorial(2) * factorial(3))

Subtract strings starting with 0 (invalid 7-digit numbers)

If first digit is 0, remaining 6 digits have one 0, two 1s, three 2s

remaining = factorial(6) // (factorial(1) * factorial(2) * factorial(3)) num_intersection = total_intersection - remaining

Inclusion-exclusion principle

union = num_A + num_B - num_intersection

num_A, num_B, num_intersection, union </PYTHON>

[THOUGHT] I need to solve the problem where we count seven-digit numbers using digits 0,1,2 (first digit not zero) such that at least one of the digits 0 or 1 appears exactly twice. I'll use inclusion-exclusion principle.

First, compute:

• |A|: exactly two � </think>

[THOUGHT]

To solve this problem, we need to count the number of valid seven-digit numbers (using digits 0, 1, 2) where at least one of the digits 0 or 1 appears exactly twice. We use inclusion-exclusion principle:

• A: Numbers with exactly two 0s
• B: Numbers with exactly two 1s
• Compute |A|, |B|, and |A ∩ B|

Step-by-step Calculation:

1. |A|: Exactly two 0s (must be in positions 2-7 since first digit can't be 0)

   • Choose 2 positions out of 6: $\binom{6}{2} = 15$
   • Remaining 5 digits (including first digit) can be 1 or 2: $2^5 = 32$
   • Total: $15 \times 32 = 480$

2. |B|: Exactly two 1s

   • Case 1: First digit is 1 → choose 1 more position from remaining 6: $\binom{6}{1} = 6$, remaining 5 digits are 0/2: $2^5 = 3 wait, no: if first digit is 1, then one more 1 in 6 positions: 6 choices. The other 5 digits (non-1) must be 0 or 2 →$ 2^5 = 32 $. So case1 =$ 6 \times 32 = 192 $
   • Case 2: First digit is 2 → both 1s in remaining 6 positions: $\binom{6}{2} = 15$, remaining 4 digits are 0/2: $2^4 = 16$ → $15 \times 16 = 240$
   • Total |B| = $192 + 240 = 432$

3. |A ∩ B|: Exactly two 0s, two 1s, three 2s

   • Total strings (including leading zero): $\frac{7!}{2! \cdot 2! \cdot 3!} = 210$
   • Invalid numbers (start with 0): Fix first digit as 0, remaining has one 0, two 1s, three 2s → $\frac{6!}{1! \cdot 2! \cdot 3!} = 60$
   • Valid: $210 - 60 = 150$

4. Final Calculation:

   $$|A \cup B| = |A| + |B| - |A \cap B| = 480 + 432 - 150 = 762$$

[PYTHON] from math import factorial, comb

Calculate |A| = exactly two 0s

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

Calculate |B| = exactly two 1s

case1 = comb(6, 1) * (2 ** 5) case2 = comb(6, 2) * (2 ** 4) num_B = case1 + case2

Calculate |A ∩ B| = exactly two 0s and two 1s

total_intersection = factorial(7) // (factorial(2) * factorial(2) * factorial(3)) remaining = factorial(6) // (factorial(1) * factorial(2) * factorial(3)) num_intersection = total_intersection - remaining

Inclusion-exclusion principle

union = num_A + num_B - num_intersection

num_A, num_B, num_intersection, union [/PYTHON]

[THOUGHT] We've computed the number of seven-digit numbers using digits 0,1,2 with at least one of the digits 0 or 1 appearing exactly twice. The result is 762.

Final answer: \boxed{762}