Pretty Easy Problem For 8-Graders

The only case in which adjacent digits differ by more than $2$ is when $1$ and $4$ are next to each other. By complementary counting, we can count all the cases that 1 and 4 are together, and subtract that from the total.

We can treat $1$ and $4$ as a "supernumber" and only $1$ digit, since they have to be together. Now, there are $3$ "digits", so the number of numbers is just $3! = 6$ . But we have to multiply this by $2$ , because we could easily get all the same numbers by switching the order of $1$ and $4$ around. Now we have $6 \times 2 = 12$ numbers that are not allowed.

Finding the total is easy: it is just $4! = 24$ . Now, the number of ways meet the standards are $24 - 12 = \fbox{12}$ .