Let be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in , but 0210222 is NOT in .Find the number of elements in such that at least one of the digits 0 and 1 appears exactly twice in .\
[THOUGHT] We need to find the number of seven-digit numbers using digits 0, 1, 2 (first digit not 0) where at least one of the digits 0 or 1 appears exactly twice. This is equivalent to the size of the union of two sets:
- A: numbers with exactly two 0's
- B: numbers with exactly two 1's
Using inclusion-exclusion: |A ∪ B| = |A| + |B| - |A ∩ B|
Calculating |A| (exactly two 0's):
- Choose 2 positions for 0's from the last 6 positions (since first digit cannot be 0): C(6,2) = 15
- The remaining 5 positions (including the first digit) are filled with 1's and 2's: 2^5 = 32
- |A| = 15 * 32 = 480
Calculating |B| (exactly two 1's):
- Total ways to choose 2 positions for 1's from 7 positions: C(7,2) = 21
- The remaining 5 positions are filled with 0's and 2's: 2^5 = 32
- This includes cases where the first digit is 0, which is invalid.
- Number of invalid cases (first digit 0): Fix first digit as 0, choose 2 positions for 1's from the last 6: C(6,2) = 15, and the remaining 4 positions: 2^4 = 16, so 15 * 16 = 240
- |B| = 21 * 32 - 240 = 672 - 240 = 432
Calculating |A ∩ B| (exactly two 0's and exactly two 1's):
- The remaining 3 digits must be 2's.
- Total arrangements of {0,0,1,1,2,2,2}: 7! / (2!2!3!) = 210
- Subtract arrangements with first digit 0: Fix first digit 0, arrange the remaining {0,1,1,2,2,2}: 6! / (1!2!3!) = 60
- |A ∩ B| = 210 - 60 = 150
Final calculation:
|A ∪ B| = 480 + 432 - 150 = 762
The number of elements in S satisfying the condition is 762. \boxed{762}
[THOUGHT] We will use Python to count the numbers that satisfy the condition by iterating over all possible seven-digit numbers formed with digits 0,1,2 (with first digit not 0) and checking the frequency of 0 and 1. [/THOUGHT] [PYTHON] import itertools
count = 0 digits = [0, 1, 2] for first in [1, 2]: # First digit cannot be 0 for last6 in itertools.product(digits, repeat=6): num_digits = [first] + list(last6) count0 = num_digits.count(0) count1 = num_digits.count(1) if count0 == 2 or count1 == 2: count += 1
print(count) [/PYTHON]
[THOUGHT] The Python tool was used and returned 762, which is the number of elements in S satisfying the condition. [/THOUGHT] \boxed{762}