Distinguishability problem / How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?
I'm not quite sure how to approach it, $\frac{3^6}{3!}$ is not an integer.  Thanks.

I'm not the best at combinatorics but here's a go. I think I remember problems like this in statistical mechanics. $B$ for box.
$B\mid\quad B\mid\quad B\mid \qquad ways$

$6\mid\quad 0\mid\quad 0\mid \qquad 1 \qquad$ all in one box.

$5\mid\quad 1\mid\quad 0\mid \qquad6\qquad$ one of the six on its own

$4\mid\quad 2\mid\quad 0\mid \qquad {6\choose2}\qquad$ choose two of the six for one box

$4\mid\quad 1\mid\quad 1\mid \qquad \frac{6*5}{2}\qquad$ pick one of the six, then one of the 5 and divide by 2 ways of doing this.

$3\mid\quad 3\mid\quad 0\mid \qquad \frac{1}{2}{6\choose3}\qquad$ six choose 3 but divide by two because the boxes are indistinguishable.

$3\mid\quad 2\mid\quad 1\mid \qquad 6*{5\choose2}\qquad$ pick one of the six then two of the five remaining

$2\mid\quad 2\mid\quad 2\mid \qquad \frac{1}{3!}{6\choose2}\cdot{4\choose2}\qquad$ two of the six, then two of the remaining four, and there are 6 ways these can be ordered so divide out by this.

Thanks to Ned and JMac31 for the help
