A Catalan number proof involving an $2$ by $n$ array












0












$begingroup$


I am struggling with this proof. Can anyone help me?



Prove that the number of distinct $2$-by-$n$ arrays



$$ begin{bmatrix}x_{11} x_{12} ... x_{1n} \ x_{21} x_{22} ... x_{2n} end{bmatrix}$$



that can be made from the numbers $1,2 ... ,2n$ such that



$$ x_{11} < x_{12} < ... < x_{1n}, $$
$$ x_{21} < x_{22} < ... < x_{2n} $$



$$x_{11} < x_{21}, x_{12} < x_{22}, ..., x_{1n} < x_{2n}, $$



equals the $n$th Catalan number $C_n$.






combinatorics catalan-numbers







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am struggling with this proof. Can anyone help me?



    Prove that the number of distinct $2$-by-$n$ arrays



    $$ begin{bmatrix}x_{11} x_{12} ... x_{1n} \ x_{21} x_{22} ... x_{2n} end{bmatrix}$$



    that can be made from the numbers $1,2 ... ,2n$ such that



    $$ x_{11} < x_{12} < ... < x_{1n}, $$
    $$ x_{21} < x_{22} < ... < x_{2n} $$



    $$x_{11} < x_{21}, x_{12} < x_{22}, ..., x_{1n} < x_{2n}, $$



    equals the $n$th Catalan number $C_n$.






    combinatorics catalan-numbers







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      I am struggling with this proof. Can anyone help me?



      Prove that the number of distinct $2$-by-$n$ arrays



      $$ begin{bmatrix}x_{11} x_{12} ... x_{1n} \ x_{21} x_{22} ... x_{2n} end{bmatrix}$$



      that can be made from the numbers $1,2 ... ,2n$ such that



      $$ x_{11} < x_{12} < ... < x_{1n}, $$
      $$ x_{21} < x_{22} < ... < x_{2n} $$



      $$x_{11} < x_{21}, x_{12} < x_{22}, ..., x_{1n} < x_{2n}, $$



      equals the $n$th Catalan number $C_n$.






      combinatorics catalan-numbers







      share|cite|improve this question











      $endgroup$




      I am struggling with this proof. Can anyone help me?



      Prove that the number of distinct $2$-by-$n$ arrays



      $$ begin{bmatrix}x_{11} x_{12} ... x_{1n} \ x_{21} x_{22} ... x_{2n} end{bmatrix}$$



      that can be made from the numbers $1,2 ... ,2n$ such that



      $$ x_{11} < x_{12} < ... < x_{1n}, $$
      $$ x_{21} < x_{22} < ... < x_{2n} $$



      $$x_{11} < x_{21}, x_{12} < x_{22}, ..., x_{1n} < x_{2n}, $$



      equals the $n$th Catalan number $C_n$.






      combinatorics catalan-numbers




      combinatorics catalan-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 '18 at 4:14









      obscurans

      1,027311




      1,027311










      asked Nov 30 '18 at 3:56









      DummKorfDummKorf

      335




      335






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Call a (possibly incomplete) arrangement of numbers in the $2 times n$ array admissible if each number is greater than all those to its left or above it. Now consider constructing a complete admissible arrangement by adding numbers $1, 2, ldots, 2n$ one by one to the empty array in increasing order. If the numbers $1, 2, ldots, m$ have been admissibly arranged, then there can be no empty cells above or to the left of the occupied cells, for in that case, putting any number from $m+1, ldots, 2n$ in the empty cells would make the arrangement inadmissible.
          Therefore at each step the lowest unassigned number can only be assigned to the leftmost empty cell in either row, or
          only in the top row if both rows contain the same number of assigned cells. In the Figure, $?$ denotes one of $1, 2, 3, 4 $ (admissibly arranged), and $*$ a possible assignment for number 5.enter image description here



          The process of adding one number at a time in the leftmost empty cell of a row can be represented as a path in a grid. Starting at $(0,0)$, when we add the new number to the top row, we move right horizontally by one unit and when we place the number in the bottom row, we move down in the grid by one unit (we represent the grid downwards). A mapping between the $2 times 5$ array and the corresponding path is shown in Figure.
          enter image description here



          Thus the number of admissible arrays is the same as the number of paths that lie above the diagonal and hence equals $C_n$.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019613%2fa-catalan-number-proof-involving-an-2-by-n-array%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Call a (possibly incomplete) arrangement of numbers in the $2 times n$ array admissible if each number is greater than all those to its left or above it. Now consider constructing a complete admissible arrangement by adding numbers $1, 2, ldots, 2n$ one by one to the empty array in increasing order. If the numbers $1, 2, ldots, m$ have been admissibly arranged, then there can be no empty cells above or to the left of the occupied cells, for in that case, putting any number from $m+1, ldots, 2n$ in the empty cells would make the arrangement inadmissible.
            Therefore at each step the lowest unassigned number can only be assigned to the leftmost empty cell in either row, or
            only in the top row if both rows contain the same number of assigned cells. In the Figure, $?$ denotes one of $1, 2, 3, 4 $ (admissibly arranged), and $*$ a possible assignment for number 5.enter image description here



            The process of adding one number at a time in the leftmost empty cell of a row can be represented as a path in a grid. Starting at $(0,0)$, when we add the new number to the top row, we move right horizontally by one unit and when we place the number in the bottom row, we move down in the grid by one unit (we represent the grid downwards). A mapping between the $2 times 5$ array and the corresponding path is shown in Figure.
            enter image description here



            Thus the number of admissible arrays is the same as the number of paths that lie above the diagonal and hence equals $C_n$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Call a (possibly incomplete) arrangement of numbers in the $2 times n$ array admissible if each number is greater than all those to its left or above it. Now consider constructing a complete admissible arrangement by adding numbers $1, 2, ldots, 2n$ one by one to the empty array in increasing order. If the numbers $1, 2, ldots, m$ have been admissibly arranged, then there can be no empty cells above or to the left of the occupied cells, for in that case, putting any number from $m+1, ldots, 2n$ in the empty cells would make the arrangement inadmissible.
              Therefore at each step the lowest unassigned number can only be assigned to the leftmost empty cell in either row, or
              only in the top row if both rows contain the same number of assigned cells. In the Figure, $?$ denotes one of $1, 2, 3, 4 $ (admissibly arranged), and $*$ a possible assignment for number 5.enter image description here



              The process of adding one number at a time in the leftmost empty cell of a row can be represented as a path in a grid. Starting at $(0,0)$, when we add the new number to the top row, we move right horizontally by one unit and when we place the number in the bottom row, we move down in the grid by one unit (we represent the grid downwards). A mapping between the $2 times 5$ array and the corresponding path is shown in Figure.
              enter image description here



              Thus the number of admissible arrays is the same as the number of paths that lie above the diagonal and hence equals $C_n$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Call a (possibly incomplete) arrangement of numbers in the $2 times n$ array admissible if each number is greater than all those to its left or above it. Now consider constructing a complete admissible arrangement by adding numbers $1, 2, ldots, 2n$ one by one to the empty array in increasing order. If the numbers $1, 2, ldots, m$ have been admissibly arranged, then there can be no empty cells above or to the left of the occupied cells, for in that case, putting any number from $m+1, ldots, 2n$ in the empty cells would make the arrangement inadmissible.
                Therefore at each step the lowest unassigned number can only be assigned to the leftmost empty cell in either row, or
                only in the top row if both rows contain the same number of assigned cells. In the Figure, $?$ denotes one of $1, 2, 3, 4 $ (admissibly arranged), and $*$ a possible assignment for number 5.enter image description here



                The process of adding one number at a time in the leftmost empty cell of a row can be represented as a path in a grid. Starting at $(0,0)$, when we add the new number to the top row, we move right horizontally by one unit and when we place the number in the bottom row, we move down in the grid by one unit (we represent the grid downwards). A mapping between the $2 times 5$ array and the corresponding path is shown in Figure.
                enter image description here



                Thus the number of admissible arrays is the same as the number of paths that lie above the diagonal and hence equals $C_n$.






                share|cite|improve this answer









                $endgroup$



                Call a (possibly incomplete) arrangement of numbers in the $2 times n$ array admissible if each number is greater than all those to its left or above it. Now consider constructing a complete admissible arrangement by adding numbers $1, 2, ldots, 2n$ one by one to the empty array in increasing order. If the numbers $1, 2, ldots, m$ have been admissibly arranged, then there can be no empty cells above or to the left of the occupied cells, for in that case, putting any number from $m+1, ldots, 2n$ in the empty cells would make the arrangement inadmissible.
                Therefore at each step the lowest unassigned number can only be assigned to the leftmost empty cell in either row, or
                only in the top row if both rows contain the same number of assigned cells. In the Figure, $?$ denotes one of $1, 2, 3, 4 $ (admissibly arranged), and $*$ a possible assignment for number 5.enter image description here



                The process of adding one number at a time in the leftmost empty cell of a row can be represented as a path in a grid. Starting at $(0,0)$, when we add the new number to the top row, we move right horizontally by one unit and when we place the number in the bottom row, we move down in the grid by one unit (we represent the grid downwards). A mapping between the $2 times 5$ array and the corresponding path is shown in Figure.
                enter image description here



                Thus the number of admissible arrays is the same as the number of paths that lie above the diagonal and hence equals $C_n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 4:45









                MuralidharanMuralidharan

                43526




                43526






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019613%2fa-catalan-number-proof-involving-an-2-by-n-array%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                    Read more

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                    Read more

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                    Image
                    1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                    Read more