Counting Problem - N unique balls in K unique buckets w/o duplication $\mid$ at least one bucket remains empty and all balls are used I am trying to figure out how many ways one can distribute $N$ unique balls in $K$ unique buckets without duplication such that all of the balls are used and at least one bucket remains empty in each distribution?
Easy, I thought. I'll just hold a bucket in reserve, distribute the balls, and place the empty bucket. I get:
$ K\cdot N! / (N-K-1)! $
Even were I sure this handles the no duplicates condition, what if $K \geq N$?
Then I get a negative factorial in the denominator. Is the solution correct and/or is there a more general solution?
Thanks!

A slightly different approach using the twelvefold way.
If $K>N$ then it doesn't matter how you distribute the balls since at least one bucket will always be empty. In this case we are simply counting functions from a $N$ element set to a $K$ element set. Therefore the number of distributions is $K^N$.
If $K=N$ then the only bad assignments are the ones in which every bucket contains precisely one ball. This happens in precisely $N!$ ways, so just subtract out these cases for a total of $N^N - N!$ distributions.
If $K < N$, we first choose a number of buckets which cannot be filled and then we fill the remaining buckets. If we choose $m$ buckets to remain empty, then the remaining $K-m$ buckets must be filled surjectively. The number of surjections for each $m$ is 
$$(K-m)!{N\brace K-m}$$ 
where the braced term is a Stirling number of the second kind. Summing over $m$ gives the required result
$$\sum_{m=1}^{K-1}(K-m)!{N\brace K-m}\binom{K}{m}$$
I am not sure if this simplifies or not. In summary, if we let $f(N,K)$ denote the number of distributions, then
$$f(N,K) = \begin{cases}K^N & K > N\\
N! & K=N\\
\sum_{m=1}^{K-1}(K-m)!{N\brace K-m}\binom{K}{m} & K < N\end{cases}$$
