Problem Analysis #16

[PYTHON] import itertools

def solve(): count = 0 # Iterate through all 7-digit tuples with elements 0, 1, 2 for p in itertools.product([0, 1, 2], repeat=7): # Check if first digit is not 0 if p[0] == 0: continue

    c0 = p.count(0)
    c1 = p.count(1)
    
    if c0 == 2 or c1 == 2:
        count += 1
        
print(count)

solve() [/PYTHON]

The number of such 7-digit numbers is calculated as follows: Let $S$ be the set of valid 7-digit numbers (first digit non-zero) using digits $\{0, 1, 2\}$. Let $A$ be the subset of $S$ where the digit 0 appears exactly twice. Let $B$ be the subset of $S$ where the digit 1 appears exactly twice. We want to find $|A \cup B| = |A| + |B| - |A \cap B|$.

1. Calculate $|A|$: The digit 0 appears exactly twice. Since the first digit cannot be 0, the two 0s must be placed in the remaining 6 positions. There are $\binom{6}{2} = 15$ ways to place the 0s. The first digit can be 1 or 2 (2 choices). The remaining 4 positions can be filled with 1 or 2 ($2^4 = 16$ choices). $|A| = 2 \times 15 \times 16 = 480$.

2. Calculate $|B|$: The digit 1 appears exactly twice.

   • Case 1: The first digit is 1. One other 1 must be placed in the remaining 6 positions ($\binom{6}{1} = 6$ ways). The other 5 positions are filled with 0 or 2 ($2^5 = 32$ choices). Count = $1 \times 6 \times 32 = 192$.
   • Case 2: The first digit is not 1. Since it cannot be 0, it must be 2 (1 choice). The two 1s are placed in the remaining 6 positions ($\binom{6}{2} = 15$ ways). The other 4 positions are filled with 0 or 2 ($2^4 = 16$ choices). Count = $1 \times 15 \times 16 = 240$. $|B| = 192 + 240 = 432$.

3. Calculate $|A \cap B|$: 0 appears twice and 1 appears twice. The remaining $7-2-2=3$ digits must be 2s. The multiset of digits is $\{0, 0, 1, 1, 2, 2, 2\}$.

   • Case 1: First digit is 1. Remaining permutations of $\{0, 0, 1, 2, 2, 2\}$ is $\frac{6!}{2!1!3!} = 60$.
   • Case 2: First digit is 2. Remaining permutations of $\{0, 0, 1, 1, 2, 2\}$ is $\frac{6!}{2!2!2!} = 90$. $|A \cap B| = 60 + 90 = 150$.

4. Final Result: $|A \cup B| = 480 + 432 - 150 = 762$.

\boxed{762}