Let $A=\{1,2,3,4,5,6\}$ How many equivalence relations on A have ... (1) ... exactly two equivalence classes of size 3? (2) ... exactly one equivalence class of size 3?
So this is what I attempted for 1, even though I have a feeling it's very wrong...
I know the sixth Bell number is 203, which represents all possible equivalence relations. I did 6choose3=20 since I wanted to find out how many ways can this set be split into 3, however for each time you split the set into 3, you leave behind an additional 3 elements, so $20\cdot2=40$ Then $203/40=5$ equivalence relations hold exactly 2 equivalence classes of size 3. I didn't even know how to attempt part 2...
The number of partitions into two sets of size three is just $\binom{6}{3}/2$
For the second part there are $\binom{6}{3}$ ways to choose the block of size $3$ and $B_3-1$ ways to partition the rest (without putting them all in the same block), so the answer is $\binom{6}{3}(B_3-1)$