75. Arrange 10 small balls labeled with numbers 1 to 10 in a row, such that the sum of the numbers on every 3 adjacent balls is a multiple of 3. The total number of arrangements is $\qquad$.
Answer: 1728.
Solution: Divide these 10 balls into 3 groups:
Group with remainder 1 when divided by 3
Group with remainder 2 when divided by 3
Group with remainder 0 when divided by 3
$(1,4,7,10)$
$(2,5,8)$;
$(3,6,9)$.

The four balls in the first group must be in the 1st, 4th, 7th, and 10th positions,
so there are
$1 \times 2 \times 3 \times 4=24$ (ways) to arrange them.

The three balls in the second group are in the 2nd, 5th, and 8th positions, with 6 ways to arrange them.

The three balls in the third group are in the 3rd, 6th, and 9th positions, with 6 ways to arrange them.
Therefore, there are
$24 \times 6 \times 6=864$ (ways) to arrange them.
Note that, the positions of the second and third groups can be swapped,
so, in total, there are
$864 \times 2=1728$ (ways) to arrange them.