Is there a combinatorial proof that the Catalan number $C_n$ satisfies $(n+1)C_n={2n choose n}$?
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I saw this question and thought that may be it is possible to prove that the $n^{text{th}}$ Catalan number $C_n$ equals $frac{1}{n+1}{2nchoose n}$ by taking a set $A$ of size $n+1$ and another set $B$ of size $C_n$ such that there exists a $1$-$1$ correspondence $Atimes Bto T$, where $T$ is the set of subsets of ${1,2,ldots,2n}$ of size $n$. I made my attempt but failed, but I am curious if there is a known bijection $Atimes Bto T$ for some $A$, $B$.
I know that there are combinatorial proofs that show $C_n={2nchoose n}-{2nchoose{n-1}}$, but I want a specific proof that shows $(n+1)C_n={2nchoose n}$. Below is my attempt.
Write $[k]={1,2,ldots,k}$. Furthermore, $binom{X}{k}$ is the set of all subsets of cardinality $k$ of a given set $X$.
Let there be $2n$ people, named, $1$, $2$, $ldots$, $2n$. The $2n$ people are seated around a round table in the counterclockwise order. Let $mathcal{P}$ denote the set of all pairings $$big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}$$ of $[2n]$ in such a way that, when $x_i$ shakes hand with $y_i$ simultaneously for every $iin[n]$, there are no crossing arms. Wlog, we assume that $x_i<y_i$ for each $iin[n]$ and that $x_1<x_2<ldots<x_n$.
Define $f:[n+1]timesmathcal{P}tobinom{[2n]}{n}$ as follows:
$$fBig(k,big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}Big)={y_1,y_2,ldots,y_{k-1},x_k,x_{k+1},ldots,x_n}$$
for each $kin[n+1]$ and $big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}inmathcal{P}$. Well, this is where my idea fails. I was hoping that $f$ will be a bijection, but it isn't even injective.
combinatorics functions reference-request combinatorial-proofs catalan-numbers
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add a comment |
$begingroup$
I saw this question and thought that may be it is possible to prove that the $n^{text{th}}$ Catalan number $C_n$ equals $frac{1}{n+1}{2nchoose n}$ by taking a set $A$ of size $n+1$ and another set $B$ of size $C_n$ such that there exists a $1$-$1$ correspondence $Atimes Bto T$, where $T$ is the set of subsets of ${1,2,ldots,2n}$ of size $n$. I made my attempt but failed, but I am curious if there is a known bijection $Atimes Bto T$ for some $A$, $B$.
I know that there are combinatorial proofs that show $C_n={2nchoose n}-{2nchoose{n-1}}$, but I want a specific proof that shows $(n+1)C_n={2nchoose n}$. Below is my attempt.
Write $[k]={1,2,ldots,k}$. Furthermore, $binom{X}{k}$ is the set of all subsets of cardinality $k$ of a given set $X$.
Let there be $2n$ people, named, $1$, $2$, $ldots$, $2n$. The $2n$ people are seated around a round table in the counterclockwise order. Let $mathcal{P}$ denote the set of all pairings $$big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}$$ of $[2n]$ in such a way that, when $x_i$ shakes hand with $y_i$ simultaneously for every $iin[n]$, there are no crossing arms. Wlog, we assume that $x_i<y_i$ for each $iin[n]$ and that $x_1<x_2<ldots<x_n$.
Define $f:[n+1]timesmathcal{P}tobinom{[2n]}{n}$ as follows:
$$fBig(k,big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}Big)={y_1,y_2,ldots,y_{k-1},x_k,x_{k+1},ldots,x_n}$$
for each $kin[n+1]$ and $big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}inmathcal{P}$. Well, this is where my idea fails. I was hoping that $f$ will be a bijection, but it isn't even injective.
combinatorics functions reference-request combinatorial-proofs catalan-numbers
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4
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See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
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– Michael Lugo
Nov 27 '18 at 15:32
add a comment |
$begingroup$
I saw this question and thought that may be it is possible to prove that the $n^{text{th}}$ Catalan number $C_n$ equals $frac{1}{n+1}{2nchoose n}$ by taking a set $A$ of size $n+1$ and another set $B$ of size $C_n$ such that there exists a $1$-$1$ correspondence $Atimes Bto T$, where $T$ is the set of subsets of ${1,2,ldots,2n}$ of size $n$. I made my attempt but failed, but I am curious if there is a known bijection $Atimes Bto T$ for some $A$, $B$.
I know that there are combinatorial proofs that show $C_n={2nchoose n}-{2nchoose{n-1}}$, but I want a specific proof that shows $(n+1)C_n={2nchoose n}$. Below is my attempt.
Write $[k]={1,2,ldots,k}$. Furthermore, $binom{X}{k}$ is the set of all subsets of cardinality $k$ of a given set $X$.
Let there be $2n$ people, named, $1$, $2$, $ldots$, $2n$. The $2n$ people are seated around a round table in the counterclockwise order. Let $mathcal{P}$ denote the set of all pairings $$big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}$$ of $[2n]$ in such a way that, when $x_i$ shakes hand with $y_i$ simultaneously for every $iin[n]$, there are no crossing arms. Wlog, we assume that $x_i<y_i$ for each $iin[n]$ and that $x_1<x_2<ldots<x_n$.
Define $f:[n+1]timesmathcal{P}tobinom{[2n]}{n}$ as follows:
$$fBig(k,big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}Big)={y_1,y_2,ldots,y_{k-1},x_k,x_{k+1},ldots,x_n}$$
for each $kin[n+1]$ and $big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}inmathcal{P}$. Well, this is where my idea fails. I was hoping that $f$ will be a bijection, but it isn't even injective.
combinatorics functions reference-request combinatorial-proofs catalan-numbers
$endgroup$
I saw this question and thought that may be it is possible to prove that the $n^{text{th}}$ Catalan number $C_n$ equals $frac{1}{n+1}{2nchoose n}$ by taking a set $A$ of size $n+1$ and another set $B$ of size $C_n$ such that there exists a $1$-$1$ correspondence $Atimes Bto T$, where $T$ is the set of subsets of ${1,2,ldots,2n}$ of size $n$. I made my attempt but failed, but I am curious if there is a known bijection $Atimes Bto T$ for some $A$, $B$.
I know that there are combinatorial proofs that show $C_n={2nchoose n}-{2nchoose{n-1}}$, but I want a specific proof that shows $(n+1)C_n={2nchoose n}$. Below is my attempt.
Write $[k]={1,2,ldots,k}$. Furthermore, $binom{X}{k}$ is the set of all subsets of cardinality $k$ of a given set $X$.
Let there be $2n$ people, named, $1$, $2$, $ldots$, $2n$. The $2n$ people are seated around a round table in the counterclockwise order. Let $mathcal{P}$ denote the set of all pairings $$big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}$$ of $[2n]$ in such a way that, when $x_i$ shakes hand with $y_i$ simultaneously for every $iin[n]$, there are no crossing arms. Wlog, we assume that $x_i<y_i$ for each $iin[n]$ and that $x_1<x_2<ldots<x_n$.
Define $f:[n+1]timesmathcal{P}tobinom{[2n]}{n}$ as follows:
$$fBig(k,big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}Big)={y_1,y_2,ldots,y_{k-1},x_k,x_{k+1},ldots,x_n}$$
for each $kin[n+1]$ and $big{{x_1,y_1},{x_2,y_2},ldots,{x_n,y_n}big}inmathcal{P}$. Well, this is where my idea fails. I was hoping that $f$ will be a bijection, but it isn't even injective.
combinatorics functions reference-request combinatorial-proofs catalan-numbers
combinatorics functions reference-request combinatorial-proofs catalan-numbers
asked Nov 27 '18 at 14:23
user614671
4
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See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
$endgroup$
– Michael Lugo
Nov 27 '18 at 15:32
add a comment |
4
$begingroup$
See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
$endgroup$
– Michael Lugo
Nov 27 '18 at 15:32
4
4
$begingroup$
See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
$endgroup$
– Michael Lugo
Nov 27 '18 at 15:32
$begingroup$
See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
$endgroup$
– Michael Lugo
Nov 27 '18 at 15:32
add a comment |
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See at the Wikipedia article on Catalan numbers (en.wikipedia.org/wiki/Catalan_number), under the sections "third proof" and "sixth proof".
$endgroup$
– Michael Lugo
Nov 27 '18 at 15:32