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Stephen

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Tea and Wine

One intuitive solution is:

The volume of liquid in both the cup and barrel remains the same as the initial state. If the net amounts of tea/wine transferred was not equal, the volumes would be different. Thus they must be equal.

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Partitioning a set

I don't think the answer you've given is correct. Suppose r = 6, k = 2. Then (6c2) * 2 / 6 = 5. But you can have: { {1,2}{3,4}{5,6} } { {1,2}{3,5}{4,6} } { {1,2}{3,6}{4,5} } { {1,3}{2,4}{5,6} } { {1,3}{2,5}{4,6} } { {1,3}{2,6}{4,5} } which isn't exhaustive but is already more than 5.

Here's another solution. There are r! permutations of the original set. Convert a permutation to a partition by treating the subsets as the first k elements, the next k elements, etc. But then there are (r/k)! * (k!)^(r/k) permutations that generate each partition -- each of the r/k subsets can be ordered in k! ways, and then the r/k subsets can be ordered in (r/k)! ways. So the number of partitions is r! / [ (r/k)! * (k!)^(r/k) ]

In the example of r = 6, k = 2, this revised formula gives 6! / (3! * (2!)^3) = 15.

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