Let $S$ be a set with $|S| = 30$ and let $\pi = \{A_i\}_{i=1}^3$ be a partition of $S$ such that each set $A_i$ of $\pi$ has ten elements. How many such partitions $\pi$ are there?
This questions seems like a deceptively easy question, i.e., just a combination problem. Pick ${30\choose 10}$ for $A_1$, ${30-10\choose 10}$ for $A_2$, and ${30-20\choose 10}$ for $A_3$ Then we have ${30\choose 10}{20\choose 10}{10\choose 10}$. But we don't care in what order the 3 partitions are so finally answer is $$\frac{{30\choose 10}{20\choose 10}{10\choose 10}} {3!}.$$ Is this a sufficient answer?
Your answer is correct. You can also write this in terms of a multinomial coefficient: $$\frac{\binom{30}{10,10,10}}{3!} = \frac{30!/10!^3}{3!}$$