Thanks.
How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?
I've tried finding the cases and then brute forcing it, but that won't work. Any help?
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Sign up to join this communityYou are looking for the ways for partitioning a set with $6$ elements in three subsets. If you allow empty subsets, the answer is given by $${6\brace 1}+{6\brace 2}+{6\brace 3} = 122$$ otherwise it is given by ${6\brace 3}=90$, with ${n\brace k}$ being a Stirling number of the second kind.