Let $p,q$ and $r$ be positive prime numbers. Calculate the number of non isomorphic abelian groups of order $p^6q^3r$.
I've tried to use the structure theorem. So we have $$G \cong \mathbb Z/\langle p^{p_1} \rangle\oplus ... \oplus \mathbb Z/\langle p^{p_l} \rangle \oplus \mathbb Z/\langle q^{q_1} \rangle \oplus ... \oplus \mathbb Z/\langle q^{q_m} \rangle \oplus \mathbb Z/\langle r\rangle $$
where $p_i, q_i \geq 1$ and $p_1+...+p_l=6$,$q_1+...+q_m=3$
I counted by hand the possibilities and it gave me $33$ groups. I was trying to think if there is an easier and nicer way using combinatorics to do this. I would appreciate if someone could tell me if my answer is correct and also if he or she has a better or more organized way of writing the solution.