# Choosing distinct balls to put into indentical urns.

"There are 10 distinct balls that can be put into two identical urns such that no urn is empty. How many ways can that be done?"

I know how to do this question when the urns are distinct (its simply 9 choose 1), but when the urns are identical the first urn containing only 1 ball and the second urn containing 9 balls is the same thing as the first containing 9 balls and the second containing 1 ball. How do I take account for all the double counting?

Also, is there a more general formula that can account for any n number of identical urns?

$10$ distinct balls can be put into $2$ distinct urns in $2^{10}$ ways.   In doing so, each such arrangement has a symmetrical case.
Hence there are $2^{10} / 2!$ distinct ways to arrange $10$ distinct balls into $2$ indistinguishable urns. (Which simplifies to $2^9$.)
Generalising this to $k$ distinct balls and $n$ indistinguishable urns gives: $\dfrac{n^k}{n!}$ distinct ways of doing so (or $\dfrac{n^{k-1}}{(n-1)!}$ if you prefer).
With the added condition that neither urn is to be empty, for $10$ balls and $2$ urns we count $\frac{2^{10}}{2!}-1$, and for $k$ balls and $n$ urns the Principle of Inclusion and Exclusion gives:$$\sum_{j=0}^{\min(n,k)-1}\frac{(-1)^j n^{k-j}}{(n-j)!}$$
• Just to add on. The number of ways to distribute $n$ distinct balls into $k$ identical boxes such that no box is empty is given by $S(n,k)$ , the Stirling numbers of the second kind. – Nicholas Oct 24 '15 at 15:03
• I have to disagree! If $n = 3$, $k = 3$, then your first formula says that there are $9/2$ ways of putting the balls into urns :( – R. Suwalski Aug 26 '17 at 8:20