Let $n\geq 4$. Prove that the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.
I am stuck on this problem but I suspect I need to establish a bijection by possibly looking at the conjugate of the Ferrer's diagram. Thanks for any help.
Proof: Let $n\geq 4$. The number of partitions of $n$ into 4 parts is a solution to the system $x_1+x_2+x_3+x_4=n$ for $x_1\geq x_2\geq x_3\geq x_4\geq 1$. Let $y_1=n-x_4$, $y_2=n-x_3$, $y_3=n-x_2$, and $y_4=n-x_1$. We know that $x_1\geq x_2\geq x_3\geq x_4\geq 1$ which implies $-x_1\leq -x_2\leq -x_3\leq -x_4\leq -1$ and thus $n-x_1\leq n-x_2\leq n-x_3\leq n-x_4\leq n-1$. So, $1\leq y_4\leq y_3\leq y_2\leq y_1\leq n-1$. Hence $y_1+y_2+y_3+y_4=4n-(x_1+x_2+x_3+x_4)=4n-n=3n$ which yields a solution to the number of partitions of $3n$ into 4 parts of size at most $n-1$. Thus there is a bijection between the system $y_1+y_2+y_3+y_4=3n$ for $n-1\geq y_4\geq y_3\geq y_2\geq y_1\geq 1$ and the system $x_1+x_2+x_3+x_4=n$ for $x_1\geq x_2\geq x_3\geq x_4\geq 1$. Therefore, the number of partitions of $n$ into 4 parts equals the number of partitions of $3n$ into 4 parts of size at most $n-1$.