Given an integer $N ≥ 0$ and an integer $K ≥ 0$, how many tuples $(n_1,\ldots,n_k)$ are there such that $n_i ≥0$ and $\Sigma n_i = N$? In other words, how many way can you "partition" $N$ into $K$ sets such that the sum of the sets adds up to $N$.
For example, $n = 5$, $k =2$ gives 6 tuples: $(5,0) , (4,1) \ldots (0,5)$.
I've come up with this recursive definition:
$P(n,k) =$ # of tuples $(n_1,..,n_k)$ such that $n_i ≥ 0$ and $\Sigma n_i = N$
$P(n,k) = \Sigma_{i=0}^n{P(i,k-1)}$
$P(0,k)=1 $
$P(n,1)=1$
A suggested answer below is that this problem is the Stirling #'s of 2nd kind but I am not sure how to reduce this problem to it. See my response below.
My question is if there's a closed form solution for $P(n,k)$?