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












0












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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







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$endgroup$

















    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






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      share|cite|improve this question













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      edited Nov 30 '18 at 4:14









      obscurans

      1,027311




      1,027311










      asked Nov 30 '18 at 3:56









      DummKorfDummKorf

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          1 Answer
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          $begingroup$

          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 $C_n$.






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            1 Answer
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            $begingroup$

            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






























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