Number of partitions of $n$ with $k$ parts equals the number of partitions of $n + \binom k {2}$ How do I prove bijectively
The number of partitions of $n$ with $k$ parts equals the number of
partitions of $n + \binom k {2}$
with $k$ distinct parts
 A: We can establish a bijection like this. 
Let us say we have a partition of $n$ into $k$ parts. Order it in non-decreasing order. Now start adding $0$ to the first part, $1$ to the second part, $2$ to the third part, $\ldots,$ $k-1$ to the $k$th part. You will get a partition of $n + \binom{k}{2}$ into $k$ distinct parts.
I hope it is clear.
A: HINT (complementary to Novice’s answer): Start with the Ferrers diagram of a partition of $n+\binom{n}2$ into $k$ distinct parts. Note that the bottom ($k$-th) row must have at least one dot, the next one up at least $2$, and so on. Now remove the following array of dots:
$$\begin{array}{c|l}
\text{Row}&\text{Remove}\\ \hline
1&\underbrace{\bullet\bullet\bullet\ldots\bullet\bullet\bullet}_{k-1}\\
2&\underbrace{\bullet\bullet\bullet\ldots\bullet\bullet}_{k-2}\\
3&\underbrace{\bullet\bullet\bullet\ldots\bullet}_{k-3}\\
&\quad\vdots\\
k-3&\bullet\bullet\bullet\\
k-2&\bullet\bullet\\
k-1&\bullet\\
k&\text{(nothing)}
\end{array}$$
How many dots have you removed? How many are left? Have you emptied any row?
