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

Hi All,

We're told that we have six marbles: 3 blue marbles, 2 red marbles, and one green marble - and that we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles CANNOT be in the same cup, meaning that we could put as many as five (all except one of the reds) in any one cup. We could leave one cup empty, or put some in each of the three cups. We're asked for the number of possible combinations. While this question might appear complex, by breaking it down into 'pieces', the individual math steps aren't too difficult.

Let's start with the green marble. It can be in any of the 3 distinct cups, so there are 3 possible placements for that marble.

Next, we know that the two red marbles CANNOT be in the same cup. If we call the cups A, B and C, then the two red marbles could be in 3 possible placements:
A&B
A&C
B&C
Thus, there are 3 possibilities for the reds

The 3 blues provide the largest number of different options. Listing them out won't be difficult though. Across the 3 cups, the 3 blues could be:
0, 0, 3
0, 3, 0
3, 0, 0
1, 1, 1
0, 1, 2
0, 2, 1
1, 0, 2
1, 2, 0
2, 0, 1
2, 1, 0
Thus, there are 10 possibilities for the blues.

Total possible arrangements = (3)(3)(10) = 90

Final Answer: A

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Rich