[BruceZ]

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There are 4^10 total ways to place the 10 black and green balls in 4 cups. We will partition the favorable outcomes into 3 cases:
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Hmm, I'm pretty sure that the answer you're getting is incorrect for several reasons:
The cases that all the green balls are in two urns include the cases where all the green balls are in one urn.

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I just realized that I had this problem and came in to fix it before I saw your post. The case where all the green balls are in three urns also includes the cases where all the green balls are in two or one urn. See my corrections in red in the original post which properly subtract these cases. This was the only type of error. The change of 0.1% makes the final answer identical to Aaron’s. Also, his counting of the number of ways to put all 4 greens in 1-3 urns are also identical to mine (4, 84, 144).


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It also appears that you're treating the urns, and balls as distinguishable for calculating the number of possible arrangements, but calculating the 'successful' cases as if the urns were indistinguishable.

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I'm treating both the balls and the urns as distinguishable for all purposes. I'm counting the ways that each distinguishable ball can select a distinguishable urn. There is no problem here.