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At here Pascal's triangle in tikz we can draw Pascal's triangle. Now I want to note some properties of coefficients C_{n+3}^4 -C_{n+2}^4-C_{n+1}^4+C_{n}^4=n^2 (the first picture) as folowing pictures

How can I draw this pictures?

I'll left to you understand the code but next is a possible solution based in Paul Gaborit's example

%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Author : Paul Gaborit (2009)

% under Creative Commons attribution license.

% Title : Pascal's triangle and Sierpinski triangle

% Note : 17 lines maximum

documentclass[border=2mm, tikz]{standalone}

%usepackage[landscape,margin=1cm]{geometry}

%pagestyle{empty}

%usepackage[T1]{fontenc}

%usepackage{lmodern}



usepackage{tikz}

usetikzlibrary{positioning,shadows,backgrounds,shapes.geometric}

begin{document}

centering



%

% x=sqrt{3/4}*minimum size

% y=3/4*minimum size

%

begin{tikzpicture}[y=7.5mm,x=8.66mm]

  % some colors

  colorlet{even}{cyan!60!black}

  colorlet{odd}{orange!100!black}

  colorlet{links}{red!70!black}

  colorlet{back}{yellow!20!white}

  % some styles

  tikzset{

    box/.style={

      regular polygon,

      regular polygon sides=6,

      minimum size=10mm,

      inner sep=0mm,

      outer sep=0mm,

      text centered,

      font=smallbfseriessffamily,

      text=#1!50!black,

      draw=#1,

      line width=.25mm,

      rotate=30,

    },

    link/.style={black,  shorten >=2mm, shorten <=2mm, line width=1mm},

  }

  % Pascal's triangle

  % row #0 => value is 1

  node[box=even] (p-0-0) at (0,0) {rotatebox{-30}{1}};

  foreach row in {1,...,11} {

     % col #0 =&gt; value is 1

    node[box=even] (p-row-0) at (-row/2,-row) {rotatebox{-30}{1}};

    pgfmathsetmacro{myvalue}{1};

    foreach col in {1,...,row} {

      % iterative formula : val = precval * (row-col+1)/col

      % (+ 0.5 to bypass rounding errors)

      pgfmathtruncatemacro{myvalue}{myvalue*((row-col+1)/col)+0.5};

      globalletmyvalue=myvalue

      % position of each value

      coordinate (pos) at (-row/2+col,-row);

      % odd color for odd value and even color for even value

      pgfmathtruncatemacro{rest}{mod(myvalue,2)}

       node[box=even] (p-row-col) at (pos) {rotatebox{-30}{myvalue}};

    }

  }

  begin{pgfonlayer}{background}

    foreach i/j in {4/1,5/1,5/2,6/2,7/6,8/6,8/7,9/7}

        node[box=even,fill=odd]  at (p-i-j) {};

  end{pgfonlayer}



    draw[link] (p-4-1.center)--(p-6-2.center);

    draw[link] (p-5-1.center)--(p-5-2.center);

    draw[link] (p-7-6.center)--(p-9-7.center);

    draw[link] (p-8-6.center)--(p-8-7.center);

    node[right=5mm of p-8-8.center, align=left]  {$(36-7)+(28-8)=49$};

    node[left=5mm of p-5-0.center, align=right]  {$(15-4)+(10-5)=16$};



end{tikzpicture}



end{document}

(19/11/18) Code updated to avoid problems mentioned in Problem with Pascal triangle example

Not a complete answer, but just intended to show a different way of drawing the triangle and also calculating the values. Requires lualatex:

documentclass[tikz,border=5]{standalone}

usetikzlibrary{shapes.geometric}

directlua{

function factorial (f)

  if f < 2 then  return 1 else  return f*factorial(f-1) end

end



function nchoosek(n, k)

  return factorial(n) / (factorial(n-k) * factorial(k))

end

}

tikzset{hexagon/.style={

  regular polygon, regular polygon sides=6, shape border rotate=30,

  minimum size=1cm, inner sep=0pt,

  draw, ultra thick, execute at begin node={setbox0hboxbgroup},

  execute at end node={egrouppgfmathparse{min(4ex,wd0)/wd0}%

    scalebox{pgfmathresult}{box0}}

}}

begin{document}

tikz[x=1cm*sin 60, y=1.5cm*cos 60]

foreach n in {0,...,11}

  foreach k in {0,...,n}    

    node [hexagon] at (-n/2+k, -n) {directlua{tex.print("" .. nchoosek(n,k))}};

end{document}

The advantage with using lualatex is that it is possible to bypass the limits on mathematical calculations in PGF (actually in TeX):