Fill $8$ boxes with $60$ items I have $8$ boxes and $60$ items: how many ways can I fill the boxes so that 


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*The order of the items in each box does not matter

*It does not matter which boxes are filled with which items. In other words $60\;0\;0\;0\;0\;0\;0\;0$ is the same as $0\;60\;0\;0\;0\;0\;0\;0$. In other words if we have a combination and we swap the items in $2$ of the boxes it will still count as one variation.

 A: I assume that the $n = 60$ items are distinct, while the $k = 8$ boxes are indistinguishable. The number of ways taking into account the order of the boxes that use $m$ specific boxes (the rest being empty) is $\binom{m}{0} m^n - \binom{m}{1} (m-1)^n + \binom{m}{2} (m-2)^n - \cdots \binom{m}{m} 0^n$ by inclusion-exclusion principle, and permuting those $m$ boxes shows that exactly $m!$ of those ways correspond to each way that uses the same $m$ boxes but where the order of the boxes is irrelevant. Thus the number of ways for the original problem that use exactly $m$ boxes is $\dbinom{n}{m} \dfrac{\sum_{i=0}^m \binom{m}{i} (m-i)^n (-1)^i}{m!}$. Then the answer can be obtained by summing that for all $m$ from $1$ to $k$.
A: To make the order of boxes irrelevant: I pick the items from 1 to 60, and put each in a box, with the restriction that if there is more than one empty box, I put it into the first one. If n boxes already contain an item, I have min (n + 1, 8) choices to put the next item. 
Let $f(n,m)$ = "Number of ways to put $n$ items into boxes, ending up with $m$ non-empty boxes". 
$f (1, 1) = 1$, and $f (1, m) = 0$ whenever $m > 1$. 
$f (n + 1, 1) = 1$, $f (n+1, m) = m f (n, m) + f (n, m-1)$ whenever $m > 1$. 
We then add $f (60, 1)$ to $f (60, 8)$. 
Using a spreadsheet, it seems to converge to $8^n / 8!$ for $n$ items into 8 boxes. Someone more clever than me can try to find a proof for this. For the question asked (60 items in 8 boxes), the result is about $1.00000089302715 * 8^{60} / 8!$.
... but it should be obvious. There are $8^{60} - 7{60}$ ways to put 60 items into exactly 8 boxes, which needs to be divided by $8!$. The number of ways to use only 7 or fewer boxes is comparatively tiny. 
