Ferrers Diagram Partition

Question

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

Already tried to do by checking transpose etc. but cannot come up with a solution.

Answer 1

Let's illustrate this situation for the case $r=3$. Initially $k=0$. Then the total is $2r+k=6$ and we seek the partitions with exactly $r+k=3$ nonzero parts. We disvover that there are three of them:

$6=4+1+1$

$6=3+2+1$

$6=2+2+2$

We translate these into Ferrers diagrams:

$4+1+1$

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$3+2+1$

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$2+2+2$

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Now increment $k$ to $1$. This increases the total to $7$ and the number of required terms to $4$. Again there are three partitions satisfying these conditions:

$7=4+1+1+1$

$7=3+2+1+1$

$7=2+2+2+1$

To get a total as small as $7$ with as many as four nonzero parts, we need all the partitions to have a $1$ at the end. We see this in the Ferrers diagrams:

$4+1+1+1$

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$3+2+1+1$

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$2+2+2+1$

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Compare the two sets of diagrams: the $k=1$ diagrams are just the $k=0$ diagram with one more single-unit row attached to the bottom.

What happens in general is that each $(r,k)$ diagram, for a total of $2r+k$ with $r+k$ nonzero parts, matches a diagram for $(r,0)$ plus each of $k$ additional units in a separate, additional row.