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Show that any number of partitions of $2r + k$ into $r + k$ parts is the same for any $k$.

I've tried this, but I haven't come up with anything; hence why I have nothing written here. But in any case, any sort of help is appreciated.

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I suppose that $r,k\in\mathbf N$, as the statement in meaningful but not true for $r,k\in\mathbf Z$ (as long as $r+k,2r+k\geq0$). – Marc van Leeuwen Mar 19 '13 at 5:56
up vote 2 down vote accepted

Make a Ferrers diagram of a partition of $2r+k$ into $r+k$ parts and remove the first column; what’s left is a partition of $r$, and every partition of $r$ can be extended uniquely to a partition of $2r+k$ by adding a column of $r+k$ dots at the left of the Ferrers diagram.

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Cool. Does this show that the number of ways to partition $n$ into $k$ parts is just the number of ways to partition $n-k$ provided $k$ is bigger than $n-k$? – muzzlator Mar 18 '13 at 22:20
@muzzlator: Yep: same argument. – Brian M. Scott Mar 18 '13 at 22:24
@BrianM.Scott Okay, so I think I get your solution, but I'm not entirely convinced (my lack of understanding). When you say every partition of $r$ can be extended uniquely, that doesn't mean it's the only way to do so right? Because I could, in some haphazard manner, just throw the $r+k$ dots on to the right side of the current partition of $r$. – AlanH Mar 18 '13 at 22:45
@Alan: I mean that the process of adding a column of $r+k$ dots at the left of the Ferrers diagram produces a unique partition of $2r+k$ into $r+k$ parts. In other words, the original map from partitions of $2r+k$ with $r+k$ parts to partitions of $r$ is reversible, and the correspondence is a bijection. – Brian M. Scott Mar 18 '13 at 22:48

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