Number of equivalence relations on a set with $kn$ elements with the condition that each equivalence class has n elements

Question

Let $ \{k,n\} \subset \Bbb{N} $. Let S be a set such that $|S| = kn$. Let M be the number of equivalence relations on $S$ such that $\forall a \in S, |\bar a| = n$. Is $M = \prod_{i=1}^{k} \binom{in}{n}$???

The reasoning that led me to conjecture that $M = \prod_{i=1}^{k} \binom{in}{n}$ is the following:

Each equivalence relation determines a partition of the set in which it is defined. So if we have $kn$ elements, there are $\binom{kn}{n}$ ways of choosing the elements in the equivalence class of a. Now we are left with $(k-1)n$ elements and there are $\binom{(k-1)n}{n}$ ways of picking the $n$ elements of the equivalence class of some $b \in S$ such that $\bar b \ne \bar a$. And repeating this until we are left with $\binom{n}{n}$ choices would result in $ \prod_{i=1}^{k} \binom{in}{n}$.

Is this correct? If it is not, then what is the number of equivalence relations on $S$ such that $\forall a \in S, |\bar a| = n$? And if it is, how could I prove it ?

Answer 1

It is not correct, but it is close. You are overcounting the number of equivalence relations by a factor $k!$ because you are counting each order of choosing the equivalence classes separately. For example, if your set is $\{1,2,3,4,5,6\}$ and you are forming equivalence classes of $2$ elements you are counting $\{\{1,2\},\{3,4\},\{5,6\}\}$ as different from $\{\{3,4\},\{1,2\},\{5,6\}\}$ and four other orders of selecting the same pairs. There are $15$ such equivalence relations.

Answer 2

I agree with your answer.

For example, if you have a total of $6$ items and you want classes of size $2$, we have $6$ choose $2$ choices for the first class and $4$ choose $2$ for the second class and $2$ choose $2$ for the third class.

That makes it a total of $15\times 6\times 1 = 90 $ such partitions.

Thus $90$ equivalence relations.