Seven-digit means the leading digit cannot be $0$. Let $A$ be the set where $0$ occurs exactly twice, and $B$ be the set where $1$ occurs exactly twice. We want $|A\cup B|=|A|+|B|-|A\cap B|$.

Count $|A|$. The two $0$s must lie among positions $2,3,\dots ,7$ (six positions), and the remaining five positions are filled by $1$ or $2$. So $|A|=\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot 2^5=15\cdot 32=480.$

Count $|B|$. Split by whether the leading digit is $1$ or not.

• Leading digit is $1$: one more $1$ goes in any of the remaining $6$ positions, and the other $5$ positions are filled with $0$ or $2$. Count $=\left(\genfrac{}{}{0ex}{}{6}{1}\right)\cdot 2^5=6\cdot 32=192.$
• Leading digit is $2$: both $1$s go in the remaining $6$ positions, and the other $4$ positions are filled with $0$ or $2$. Count $=\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot 2^4=15\cdot 16=240.$

So $|B|=192+240=432.$

Count $|A\cap B|$. Two $0$s, two $1$s, and three $2$s, with leading digit not $0$. Choose $2$ slots out of the last $6$ for the zeros: $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$. From the remaining $5$ slots, choose $2$ for the ones: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. The other $3$ slots are forced to be $2$. Total $=15\cdot 10=150.$

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

$\therefore$ The answer is $\boxed{762}$.