Set Partition of N elements into K groups. All equal elements.

Question

I am trying to get the total number of possible set partitions of N elements into K groups, in a case where all the elements are the same. For example, in the case of $N=2$ and $K=3$ the possible partitions are 6 and would be:

(the numbers are the total of elements in each group)

2 0 0

1 1 0

1 0 1

0 1 1

0 2 0

0 0 2

The case for $N=3$ and $K=3$ seems to result in 10.

I know it is a sum that depends on how many elements were put in the "first" group (it starts with $K+...$, which is the case where all the elements went into one group) , but I can't grasp what is the exact pattern here. This looks like a special case of the Bell number I think, but I wasn't able to use that either.

Answer 1

Think of your elements as of $n$ balls which are to be placed into $k$ urns. We can place the balls in a row, and then insert $k-1$ dividors, separating thge balls into $k$ groups. Each group containing the balls which are to be placed into the corresponding urn. We have $n+k-1$ places for dividors and balls, and k-1 dividors to choose from them. So there are: $\binom{n+k-1}{k-1}$ possibilities.