Suppose we have six marbles: 3 blue marbles, 2 red marbles,

3 blue marbles:
Use the SEPARATOR METHOD -- a great way to count the number of ways n identical objects can be distributed among r distinct containers.
Here, 3 identical blue marbles are to be distributed among -- at most -- 3 distinct cups.
Thus, we need 3 marbles and 2 separators:
O|O|O

Any arrangement of the elements above represents one way to distribute the 3 identical marbles among 3 cups B, W and P.
O|O|O = B gets 1 marble, W gets 1 marble, P gets 1 marble.
OO||O = B gets 2 marbles, W gets 0 marbles, P gets 1 marble.
OOO|| = B gets all 3 marbles.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 5 elements above (the 3 identical marbles and the 2 identical separators).
The number of ways to arrange 5 elements = 5!.
But when an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by the number of ways to arrange the 3 identical marbles (3!) and the number of ways to arrange the 2 identical separators (2!):
5!/(3!2!) = 10.

2 red marbles:
Since the 2 red marbles cannot be in the same cup, we must choose 2 cups to contain 1 red marble each.
From 3 cups, the number of ways to choose 2 = 3C2 = (3*2)/(2*1) = 3.

Green marble:
Number of options for the 1 green marble = 3. (Any of the 3 cups.)

To combine the options for each colored marble, we multiply:
10*3*3 = 90.

The correct answer is A.