25. There are 3 red, 3 blue, and 3 green chips, a total of 9 chips of the same shape, to be placed in a $3 \times 3$ grid such that no two chips of the same color are adjacent. How many different ways can this be done?
(A) 12 (B) 18 (C) 24 (D) 30 (E) 36
25. E.

Consider two cases.
(1) Take one red, one blue, and one green chip to place in the first row, there are $3!=6$ different ways to place them.

Then take one red, one blue, and one green chip to place in the second row, considering that they should not be adjacent vertically to the chips in the first row, there are 2 different ways to place them.

Finally, take one red, one blue, and one green chip to place in the third row, similarly, there are 2 different ways to place them.
Thus, there are 3! $\times 2 \times 2=24$ different ways to place them.
(2) If there are at most two chips of the same color in each row, the first row has 3 ways to place the chips, the second row has 2 ways, and the third row has 2 ways.
Thus, there are $3 \times 2 \times 2=12$ different ways to place them.
Therefore, the total number of ways is $N=24+12=36$.