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I'm not sure what I'm missing. I think I'm thinking too hard about finding this bijection. Please help!

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2 Answers 2

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 $n-1$.

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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.)

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