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.
 A: This problem can be solved in general with Stirling numbers of the second kind. For more on general distribution problems see here.
In your case you will want to sum $S(6,1) + S(6,2) + S(6,3)$.
A: Hint:
$\frac{3^6}{3!}$ would work if the contents of the three boxes were always distinguishable. There is a problem when two of the boxes are empty. In other words, there are three ways—not six—to put all the balls in one of the boxes.

Solution:
Since there are $3$ ways to put all the balls in one box, it follows that there are $3^n-3$ ways to put the balls in boxes where at most one of the boxes is empty. In this latter case, we can divide by 3! since the three boxes along with their contents are now distinguishable.
Counting these, then adding back the one way to put all the balls in one box, gives the following:
$$\frac{3^6-3}{3!} + 1 = 122.$$
So you were close! Incidentally, if you get into Sterling numbers of the second kind, $S(n,k)$, you can check that for all $n$, we indeed have $S(n,1)+S(n,2)+S(n,3) = \dfrac{3^n-3}{3!} + 1$.
A: 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
