How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

Question

anonymous · Answer

Would you believe me if I said
$$\huge\sum_{i=0}^32^{6-i}\binom 3i=216~?$$

anonymous · Answer

Hmm that might not be right. I think it considers the boxes to be distinguishable... Ugh

anonymous · Answer

$\dbinom3i$ counts the numbers of ways you can assign $i=0,1,...,5,6$ balls in the first box, and $\large 2^{6-i}$ counts the ways the other two boxes can hold the remaining $6-i$ balls. I think the formula double counts in some places... I'll have to look this over later.