Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Using Ferrer's diagram, prove that the number of partitions of n in which each part is 1 or 2 is equal to the number of partitions of n+3 which has exactly two distinct parts.

Any help please, all I can find is The number of partitions of n with number of parts at most k is equal to the number of partitions of n + k with number of parts exactly k. However for here I can't have k to equal to 2 and 3. Please any ideas.

share|cite|improve this question
Seems to me all you need is to show that the number of partitions of $n+2$ into $2$ parts equals the number of partitions of $n+3$ into $2$ distinct parts. You may have to treat separately the cases $n$ even and $n$ odd. – Gerry Myerson May 9 '13 at 12:18

If you start with a partition of $n$ in which each part has size $1$ or $2$ and take its conjugate, you get a partition of $n$ into one or two parts. The only time you get just one part, however, is when you started with the partition of $n$ into $n$ parts of size $1$. And the only time you get two identical parts is when $n$ is even, and you start with the partition of $n$ into $n/2$ parts of size $2$.

How can you add three dots to the Ferrers diagram of the original partition in such a way that the conjugate has two distinct dots, and the original partition is uniquely recoverable from the conjugate?

If $\lambda$ is the original partition, $\mu$ is the partition of $n+3$ after you’ve added the $3$ dots, and $\mu'$ is the conjugate of $\mu$, you must make sure that $\mu$ has only parts of size $1$ and $2$ and has at least one part of size $2$; that will ensure that $\mu'$ has two parts. How can you add the $3$ dots to ensure that $\mu'$ has two distinct parts? What does that say about $\mu$?

share|cite|improve this answer

You can see it using the Ferrer's diagram of the partition of $n$ into 1's and 2's : $$\begin{array}{cc}\bullet&\\\bullet&\\\bullet&\bullet\\\bullet&\bullet\\\bullet&\bullet\end{array}\tag{8=1+1+2+2+2}$$ and turn it like this $$\begin{array}{ccccc}\bullet&\bullet&\bullet&&\\\bullet&\bullet&\bullet&\bullet&\bullet\end{array}\tag{8=3+5}$$ add one bullet to the first row (it can't be empty) and two to the second row because you must avoid the equality case $$\begin{array}{ccccccc}\bullet&\bullet&\bullet&\color{red}\bullet&&&\\\bullet&\bullet&\bullet&\bullet&\bullet&\color{red}\bullet&\color{red}{\bullet}\end{array}\tag{11=4+7}$$ The reciprocal is found by similar manipulations.

Using these diagrams helps to find the demonstration in terms of just numbers: Suppose you have a partition of $n$ that contains only 1's and 2's. You can count the number of 1's and 2's and call them $p\geq0$ and $q\geq0$ respectively. Thus you have $p+2q=n$. Write $p'=q+1$, $q'=p+q+2>p'$: you have found a partition $n+3=p'+q'$ into two distinct parts.

Reciprocally, suppose you have $n+3=s+t$ with $s>t\geq1$, then write $s'=s-t-1\geq0$, $t'=t-1\geq0$ and you have $n=2s'+t'$.

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.