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(D) Let $A$ be the set of numbers where digit $0$ appears exactly twice. Let $B$ be the set of numbers where digit $1$ appears exactly twice. We want

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$762$

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(D) Let $A$ be the set of numbers where digit $0$ appears exactly twice. Let $B$ be the set of numbers where digit $1$ appears exactly twice. We want to find $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.$1$. Calculation of $n(A)$:The first digit cannot be $0$. So,we choose $2$ positions for $0$ out of the remaining $6$ positions in $\binom{6}{2}$ ways. The remaining $5$ positions can be filled by $1$ or $2$ in $2^5$ ways. Thus,$n(A) = \binom{6}{2} 	imes 2^5 = 15 	imes 32 = 480$.$2$. Calculation of $n(B)$:Case $I$: $1$ is at the first position. We need one more $1$ in the remaining $6$ positions,which can be placed in $\binom{6}{1}$ ways. The remaining $5$ positions can be filled by $0$ or $2$ in $2^5$ ways. Number of ways $= 6 	imes 32 = 192$.Case $II$: $1$ is not at the first position. The first position can be $2$ (only $1$ option,as it cannot be $0$). We choose $2$ positions for $1$ out of the remaining $6$ positions in $\binom{6}{2}$ ways. The remaining $4$ positions can be filled by $0$ or $2$ in $2^4$ ways. Number of ways $= 15 	imes 16 = 240$.So,$n(B) = 192 + 240 = 432$.$3$. Calculation of $n(A \cap B)$:We need exactly two $0$s and exactly two $1$s. The first digit cannot be $0$. If the first digit is $1$,we need one more $1$ in $\binom{6}{1}$ ways and two $0$s in $\binom{5}{2}$ ways. The remaining $3$ positions can be filled by $2$ in $1$ way. Ways $= 6 	imes 10 = 60$.If the first digit is $2$,we need two $0$s in $\binom{6}{2}$ ways and two $1$s in $\binom{4}{2}$ ways. The remaining $2$ positions can be filled by $2$ in $1$ way. Ways $= 15 	imes 6 = 90$.So,$n(A \cap B) = 60 + 90 = 150$.$4$. Final result:$n(A \cup B) = 480 + 432 - 150 = 762$.

Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example,$2210222$ is in $S$,but $0210222$ is not in $S$. Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$,is equal to $....$

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