Question 10

Let $A$ be the set of numbers where digit $0$ appears exactly twice, and $B$ be the set where digit $1$ appears exactly twice. We need to find $|A \cup B| = |A| + |B| - |A \cap B|$.

1. Finding $|A|$: Digit $0$ cannot be at the 1st position. Choose $2$ positions for $0$ out of the remaining $6$: $\binom{6}{2} = 15$ ways. The 1st position can be filled by $1$ or $2$ ($2$ ways). The remaining $4$ positions can be filled by $1$ or $2$ ($2^4 = 16$ ways). $|A| = 15 \times 2 \times 16 = 480$.

2. Finding $|B|$: Case i: 1st digit is $1$. Remaining $6$ positions must contain exactly one $1$. $\binom{6}{1} = 6$ ways. Other $5$ positions filled by $0$ or $2$: $2^5 = 32$ ways. Total = $6 \times 32 = 192$. Case ii: 1st digit is $2$. Remaining $6$ positions must contain exactly two $1$s. $\binom{6}{2} = 15$ ways. Other $4$ positions filled by $0$ or $2$: $2^4 = 16$ ways. Total = $15 \times 16 = 240$. $|B| = 192 + 240 = 432$.

3. Finding $|A \cap B|$: Case i: 1st digit is $1$. One more $1$ needed in remaining $6$ slots, and two $0$s needed. $\binom{6}{1} \times \binom{5}{2} = 6 \times 10 = 60$. Remaining $3$ slots must be $2$. Case ii: 1st digit is $2$. Two $1$s and two $0$s needed in remaining $6$ slots. $\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90$. Remaining $2$ slots must be $2$. $|A \cap B| = 60 + 90 = 150$.

4. Calculation: $|A \cup B| = 480 + 432 - 150 = 912 - 150 = 762$. Final Answer: 762