How can I draw Pascal's triangle with some its properties?

Question

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?

Answer 1

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=\small\bfseries\sffamily,
      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};
      \global\let\myvalue=\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

Answer 2

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={\setbox0\hbox\bgroup},
  execute at end node={\egroup\pgfmathparse{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):