Question 30Pia places the integers 1 to 9 in nine boxes in a row, one in each, so any three adjacent boxes have a sum divisible by 3. How many arrangements are possible? $6^4$$6^3$$2^9$6 × 5 × 4 × 3 × 2 × 19 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1