How many ways are there to divide elements into equal unlabelled sets? How many ways are there to divide N elements into K sets where each set has exactly N/K elements.
For the example of 6 elements into 3 sets each with 2 elements. I started by selecting the elements that would go in the first set (6 choose 2) and then those that would go into the second as (4 choose 2) and then the 2 remaining elements into the third set. This gives, (6 choose 2) * (4 choose 2). In general 
(N choose N/K) * (N-(N/K) choose N/K) * (N-(2*N/K) choose N/K) * ... * 1


In response to your comment on Ross's answer, there is a more compact form:
$$
\text{number of partitions }=P_{n,k}=\frac{n!}{(n/k)!^{k}k!}
$$
To prove this, we instead prove
$$
n!=P_{n,k}\cdot (n/k)!^k\cdot k!
$$
To prove that, ask how many ways are there permute $n$ elements? There are certainly $n!$ such ways. But another to form a permutations is this: first, divide the $n$ elements into $k$ equal sets $(P_{n,k}$ ways to do this). Then, permute each element within each of the sets (they all have size $n/k$, and there are $k$ of them, so there are $(n/k)!^k$ ways. Finally, put the $k$ sets themselves into some order $(k!$ ways). 
You can check that $P_{6,3}=\frac{6!}{2!^3\cdot3!}=15$, and that indeed there are $5$ choices for what the set containing $1$ is, and then $\binom{4}2=3$ ways to partition the remaining 4 elements into sets.
