TITLE: Partition Bijection
QUESTION [5 upvotes]: I'm not sure what I'm missing.  I think I'm thinking too hard about finding this bijection.  Please help!

REPLY [5 votes]: Any partition of $n$ into $4$ parts must have each part no bigger than $n-1$.  But
$$a+b+c+d=3n \ \ \ \Leftrightarrow \ \ \ (n-a)+(n-b)+(n-c)+(n-d)=n,$$
so the number of partitions of $3n$ into $4$ parts, each between $1$ and $n-1$, equals the number of partitions of $n$ into $4$ parts, each between $1$ and $n-1$.

REPLY [3 votes]: Start with the Ferrers diagram of the partition of $n$. Pad each row with enough dots to make a $4\times n$ array of dots. Remove the original Ferrers diagram, and rotate the array of added dots $180^\circ$; the result is the Ferrers diagram of a partition of $3n$ into $4$ parts, each of size at most $n-1$. Verify that all such partitions of $3n$ are obtained in this way.
(This is the visual version of David Moews’s argument.)