0 above the Pascal-triangle?

Question

The Pascal triangle for binomial coefficients starts with 1 in line 1 and goes on with 1 1 in line 2, 1 2 1 in line 3, and so on.

The answers to this question

Pascal's triangle in tikz

show how one may write up solutions in tikz.

Here is the adapted version which I prefer:

\begin{tikzpicture}[rotate=-90]
\foreach \x in {0,1,...,3}
{
    \foreach \y in {0,...,\x}
    {
        \pgfmathsetmacro\binom{factorial(\x)/(factorial(\y)*factorial(\x-\y))}
        \pgfmathsetmacro\shift{\x/2}
        \node[xshift=-\shift cm] at (\x,\y) {\pgfmathprintnumber\binom};
    }
}
\end{tikzpicture}

How do I write up a variant which has 0 in the first line, 1 in the second line, 1 1 in the third line and thereafter goes on as the usual Pascal's triangle?

Answer 1

You can just add

\node at (-1,0) {0};

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}[rotate=-90]
\node at (-1,0) {0};
\foreach \x in {0,1,...,3}
{
    \foreach \y in {0,...,\x}
    {
        \pgfmathsetmacro\binom{factorial(\x)/(factorial(\y)*factorial(\x-\y))}
        \pgfmathsetmacro\shift{\x/2}
        \node[xshift=-\shift cm] at (\x,\y) {\pgfmathprintnumber\binom};
    }
}
\end{tikzpicture}

\end{document}
    

Answer 2

Just a small alternative to @Torbjorn's answer (a trick he taught me here ) to avoid using (annoying IMHO, \pgfmathsetmacro), include the variables to evaluate directly in the loop.

More details in pfg manual v3.1.5.b p1003, section 89 Repeating Things: The Foreach Statement.

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}[rotate=-90]
\node at (-1,0) {0};
\foreach \x in {0,1,...,3}
{
 \foreach [evaluate ={
           \binom = factorial(\x)/(factorial(\y)*factorial(\x-\y));
           \shift = \x/2 ;
           }] \y in {0,...,\x}
    {\node[xshift=-\shift cm] at (\x,\y) {\pgfmathprintnumber\binom};}
}
\end{tikzpicture}

\end{document}