Prove there are $75$ ways to put five different fish in three buckets s.t. two buckets are identical and no bucket is empty

Question

Let $\{\}_1, \{\}_2, \{\}_3$ stand for three distinct buckets. The number of onto functions is given by $n!S(k, n)$ where $S(k, n)$ is a Stirling Number of the Second Kind. So, $3!S(5, 3)$ is a number of ways to distribute five distinct fish into three different buckets with none empty. But this number counts situations like $\{\text{bass}\}_1, \{\text{trout, salmon}\}_2, \{\text{sardine, herring}\}_3$ and $\{\text{trout, salmon}\}_1, \{\text{bass}\}_2, \{\text{sardine, herring}\}_3$ as distinct even though two out of three buckets are identical. We need to divide out the redundant permutation meaning the final answer looks like $\displaystyle{\frac{3!S(5, 3)}{2!} = 75}$. Does it make sense to you? Thanks.