Problem 1067

Solution:

Removing the string "2020", there are two $0$'s, four $1$'s, and two $2$'s remaining. There are $\frac{8!}{2!4!2!} = 420$ ways to arrange these digits, multiplied to $9$ possible placements for the string "2020", for a product of $3780$.

However, by PIE, we still need to subtract the number of arrangements with string "202020" and add back the number of arrangements with string "20202020", each of which account for $7 \cdot \frac{6!}{1!4!1!} = 210$ and $5 \cdot \frac{4!}{0!4!0!} = 5$ arrangements, respectively. This gives us a total of $3780 - 210 + 5 = 3575$ possible permutations.