Let X be a set with 30 elements. How many equivalence relations can be defined on X if every class of the relation must have exactly 6 elements? I am not really sure about my approach. I'm thinking of the ways to partition the set X into 5 classes with 6 elements each.
I am thinking of the "word" $a_1a_2a_3a_4a_5a_6|a_7\dots a_{30}$ that has $30$  $a_i$ elements and 4 bars. I am thinking of the permutations of this word with fixed bars, and dividing the total of permutations by 6! repeated cases I'd get since every partition is appearing 6! times. But permutating with fixed bars is the same as them not being there at all, so the answer is the total number of permutations divided by 6!
$Var(30,24) = \frac{30!}{6!}$
Is this correct? Is there another way to think this problem?
 A: There are $30\choose 6$ ways to pick the first class, then $24\choose 6$ to choose the second class, then $18\choose 6$, $12\choose 6$, and $6\choose 6$. However, this overcounts by a factor of $5!$ because the order of picking the equivalence classes does not matter. So we end up with
$$ \frac{30!}{6!24!}\cdot \frac{24!}{6!18!}\cdot\frac{18!}{6!12!}\cdot\frac{12!}{6!6!}\cdot\frac{6!}{6!0!}\cdot\frac1{5!}=\frac{30!}{6!^55!}$$

Or: There are $29\choose 5$ ways to extend the smallest element to an equivalence class. After that, there are $23\choose 5$ ways to extend the then smallest unused element to an equivalence class, then $17\choose 5$, $11\choose 5$, and $5\choose 5$. Now we get
$$ \frac{29!}{5!24!}\cdot \frac{23!}{5!18!}\cdot \frac{17!}{5!12!}\cdot \frac{11!}{5!6!}\cdot \frac{5!}{5!0!}.$$
Can you see wha that's the same?
A: I always like to visually think about it as follows:
Line up all $30$ elements. Clearly there are $30!$ to do this.
Now create the $5$ groups by putting the first $6$ elements of the line-up in a group, the second $6$ in another, etc.
In other words, each of the $30!$ possible line-ups translates into a partition.
However, this overcounts the number of possible partitions in two ways:
First, we can shuffle around the first $6$ elements in the line-up, and still end up with the same partition. Of course, the same goes for the second $6$ elements, etc. So, we need to divide by $6!^5$
Second, we can swap any of the groups of $6$ elements forming a group with any other $6$ elements forming a group. That is, the are $5!$ ways in which these groups will be ordered in the loin-up, each resulting in the same partition. So, we also need to divide by $5!$
Total:
$$\frac{30!}{(6!)^5 \cdot 5!}$$
