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I suppose that this question has already been asked, but I couldn't find it. Suppose we have a set $A$ with $nk$ elements. How many partitions of this set into sets of k elements are there?. For $n=k=2$, there are 3:
$\{\{\{1,2\}, \{3,4\}\}, \{\{1,3\}, \{2,4\}\}, \{\{1,4\},\{2,3\}\}$
NOTE: The case $n=2$ is solved here: Counting the number of partitions having blocks of cardinality 2 and non-distinct elements
One way to solve it: first put all $nk$ items in order: there are $(nk)!$ ways. Now chop them into blocks of $k$ and you have a partition. Each block can be reordered in $k!$ ways, then the blocks can be put in order in $n!$ ways. In all, we have $\frac{(nk)!}{(k!)^nn!}$ ways to make the partition.
Another way is to say there are ${nk \choose k}$ ways to get the first partition, ${(n-1)k \choose k}$ to get the second, and so on. Again you can choose the partitions in $n!$ orders. This gets you the same answer.
