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Suppose that you have a set of $12$ colored balls, two each of six different colors $C_1$ through $C_6$. Find the number of six-ball combinations if balls of the same color are considered identical.

I think the problem quoted above is identical to the one in this question. In both problems we are counting combinations of length six with repetition. Each combination in both problems contains at most two copies of the same element, so the answer to my problem must be $141$ as well. Is this correct?

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Your reasoning is correct, the answer is indeed 141. However, I would solve this problem in a more straightforward way. Since every color can occur at most two times, one should simply consider the following cases:

  1. Combinations of $(1, 1, 1, 1, 1, 1)$: ${6 \choose 6} = 1$ option;

  2. Combinations of $(2, 1, 1, 1, 1)$: ${6 \choose 1} {5 \choose 4} = 30$ options;

  3. Combinations of $(2, 2, 1, 1)$: ${6 \choose 2} {4 \choose 2} = 90$ options;

  4. Combinations of $(2, 2, 2)$: ${6 \choose 3} = 20$ options.

As such, there are $1 + 30 + 90 + 20 = 141$ valid combinations.

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