11.5. Find the number of different arrangements in a row of all natural numbers from 1 to 10 such that the sum of any three consecutive numbers is divisible by 3.
Answer: $4! \cdot 2 \cdot 3! \cdot 3! = 1728$

Solution. From the condition, it follows that the remainders of the numbers standing two apart when divided by 3 are equal. Therefore, the numbers standing at positions 1, 4, 7, and 10, as well as those at positions 2, 5, and 8, and at positions 3, 6, and 9, have equal remainders when divided by 3. Among the numbers from 1 to 10, four identical remainders of 1 are given only by 1, 4, 7, and 10, so any permutation of these numbers can be placed at positions 1, 4, 7, and 10, giving a total of 4! options. Similarly, at positions 2, 5, and 8, any arrangement of the numbers 2, 5, 8, or the numbers 3, 6, and 9 can be placed, and at positions 3, 6, and 9, the opposite. In total, we get $4! \cdot 3! \cdot 3! \cdot 2$ options.