Combinatorics problem based on Ferrers graph
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Need help with this proof using Ferrers' graph or otherwise.
Show that the number of partitions of $r+k$ into $k$ parts is equal to
- The number of partitions of $r + {k+1 choose 2}$ into $ k $ distinct parts
- The number of partitions of $r$ into parts of size at most $k$
combinatorics integer-partitions
$endgroup$
add a comment |
$begingroup$
Need help with this proof using Ferrers' graph or otherwise.
Show that the number of partitions of $r+k$ into $k$ parts is equal to
- The number of partitions of $r + {k+1 choose 2}$ into $ k $ distinct parts
- The number of partitions of $r$ into parts of size at most $k$
combinatorics integer-partitions
$endgroup$
add a comment |
$begingroup$
Need help with this proof using Ferrers' graph or otherwise.
Show that the number of partitions of $r+k$ into $k$ parts is equal to
- The number of partitions of $r + {k+1 choose 2}$ into $ k $ distinct parts
- The number of partitions of $r$ into parts of size at most $k$
combinatorics integer-partitions
$endgroup$
Need help with this proof using Ferrers' graph or otherwise.
Show that the number of partitions of $r+k$ into $k$ parts is equal to
- The number of partitions of $r + {k+1 choose 2}$ into $ k $ distinct parts
- The number of partitions of $r$ into parts of size at most $k$
combinatorics integer-partitions
combinatorics integer-partitions
edited Oct 12 '12 at 21:14
MJD
47.4k29214396
47.4k29214396
asked Oct 12 '12 at 20:24
NeelNeel
61
61
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1 Answer
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$begingroup$
(i) Take the diagram of an arbitrary partition of $r+k$ into $k$ parts, then add $0$ to the smallest part, add $1$ to the second smallest part, add $2$ to the third, $ldots$ add $(k-1)$ to the largest part and what have you got?
(ii) Take the diagram of an arbitrary partition of of $r+k$ into $k$ parts, then subtract $1$ from each part and what have you got? Now reflect it so rows become columns and columns become rows and what have you got?
$endgroup$
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
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1 Answer
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1 Answer
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$begingroup$
(i) Take the diagram of an arbitrary partition of $r+k$ into $k$ parts, then add $0$ to the smallest part, add $1$ to the second smallest part, add $2$ to the third, $ldots$ add $(k-1)$ to the largest part and what have you got?
(ii) Take the diagram of an arbitrary partition of of $r+k$ into $k$ parts, then subtract $1$ from each part and what have you got? Now reflect it so rows become columns and columns become rows and what have you got?
$endgroup$
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
add a comment |
$begingroup$
(i) Take the diagram of an arbitrary partition of $r+k$ into $k$ parts, then add $0$ to the smallest part, add $1$ to the second smallest part, add $2$ to the third, $ldots$ add $(k-1)$ to the largest part and what have you got?
(ii) Take the diagram of an arbitrary partition of of $r+k$ into $k$ parts, then subtract $1$ from each part and what have you got? Now reflect it so rows become columns and columns become rows and what have you got?
$endgroup$
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
add a comment |
$begingroup$
(i) Take the diagram of an arbitrary partition of $r+k$ into $k$ parts, then add $0$ to the smallest part, add $1$ to the second smallest part, add $2$ to the third, $ldots$ add $(k-1)$ to the largest part and what have you got?
(ii) Take the diagram of an arbitrary partition of of $r+k$ into $k$ parts, then subtract $1$ from each part and what have you got? Now reflect it so rows become columns and columns become rows and what have you got?
$endgroup$
(i) Take the diagram of an arbitrary partition of $r+k$ into $k$ parts, then add $0$ to the smallest part, add $1$ to the second smallest part, add $2$ to the third, $ldots$ add $(k-1)$ to the largest part and what have you got?
(ii) Take the diagram of an arbitrary partition of of $r+k$ into $k$ parts, then subtract $1$ from each part and what have you got? Now reflect it so rows become columns and columns become rows and what have you got?
answered Oct 12 '12 at 20:46
HenryHenry
100k481167
100k481167
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
add a comment |
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
$begingroup$
Very nice hint.
$endgroup$
– Brian M. Scott
Oct 13 '12 at 4:11
add a comment |
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