Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

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

share|cite|improve this question

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 $n-1$.

share|cite|improve this answer

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.