How can I solve this partitions question?

Question

Let $p_n$ denote the number of partitions of $n$ in which no part occurs more than twice and let $q_n$ denote the number of partitions of $n$ in which no part is divisible by $3$. Show that $p_n=q_n$, $n≥0$.

Answer 1

No part used more than twice ... $(1+x^n+x^{2n})$ \begin{eqnarray*} \prod_{n=1}^{\infty} (1+x^n+x^{2n}) \end{eqnarray*} Now multiply top & bottom by $(1-x^n)$ ... cancel $(1-x^{3n})$ ...

Answer 2

Let $P_1(n)$ be the set of partitions of $n$ that have a part that occurs more than twice but do not have a part that is a multiple of $3$. Let $P_2(n)$ be the set of partitions of $n$ that do not have a part that occurs more than twice but do have a part that is a multiple of $3$. Finally, let $P_3(n)$ be the set of partitions of $n$ that do not have a part that occurs more than twice and do not have a part that is a multiple of $3$. You want to show that $|P_1(n)\cup P_3(n)|=|P_2(n)\cup P_3(n)|$, and because the sets $P_1(n),P_2(n)$, and $P_3(n)$ are pairwise disjoint, this is true if and only if $|P_1(n)|=|P_2(n)|$. Thus, you can prove the result by showing that there is a bijection between $P_1(n)$ and $P_2(n)$.

HINT: There is a very easy way to convert a part that is a multiple of $3$ into a part that occurs more than twice and vice versa.