Can someone lead me to a method for calculating the number of sequences of length $n$ if the terms of the sequence are chosen (with replacement) from a set with $k$-elements under the condition that all $k$-elements are chosen at least once? It's not my field.


I lead myself to consider the recurrence relation 
$$ R(k,n) = k R(k,n-1) + k R(k-1,n-1) $$
where $R(k,k) = k!, R(k,n) = 0$ if $n \lt k$ and $R(1,n) = 1$ for all $n$, but struggled to close it out. Is there a method for this?


$R(n,k)$ is the number I want, so from my point of view I'm asking one question. From your point of view I'm sure I'm asking two. Many Thanks.



Counting sequences - a recurrence relation.
These are related to Stirling numbers. These don't have a "closed form".
Your $R(k,n)=k!S(n,k)$ where $S(n,k)$ is the Stirling number of
the second kind as defined at
<http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind> .


