Answer: D

Consider how many different advertisements there are for each of the six sequential spots. Our six advertisements are X1, X2, X3, A, B, and C. X1, X2, and X3 must go in spots 2, 4, and 6, while the other three go in spots 1, 3, and 5. Here, then, are the number of possible ads for each of the six spots:
1: 3 possibilities: A, B, or C.
2: 3 possibilities: X1, X2, or X3
3: 2 possibilities: A, B, or C, but whichever ad was placed in spot 1 cannot be chosen again
4: 2 possibilities: X1, X2, or X3, but whichever ad was used in spot 2 cannot be chosen again
5: 1 possibility: whichever is remaining of A, B, and C
6: 1 possibility: whichever is remaining of X1, X2, and X3
The number of possible arrangements is the product of those six numbers:
(3)(3)(2)(2)(1)(1) = 36, choice (D).