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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.

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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
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2 Answers

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 $\mu$?

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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 $n=2s'+t'$.

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