In de driehoek van Pascal is elk getal de som van de twee getallen die er schuin boven staan.

$
\left( {\begin{array}{*{20}c}
n\\
{k-1}\\
\end{array}} \right) + \left( {\begin{array}{*{20}c}
n\\
k\\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{n+1}\\
k\\
\end{array}} \right)
$

Bewijs

$
\begin{gathered}
  \left. \begin{gathered}
  \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!}}
{{k!\, \cdot \left( {n - k} \right)!}}\, \\
  \,\left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) = \frac{{n!}}
{{(k - 1)!\, \cdot \left( {n - (k - 1)} \right)!}} \\
\end{gathered}  \right\} \to \left( {\begin{array}{*{20}c}
   {n + 1}  \\
   k  \\

 \end{array} } \right) = \frac{{(n + 1)!}}
{{k!\, \cdot (n + 1 - k)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!}}
{{(k - 1)!\, \cdot \left( {n - (k - 1)} \right)!}} + \frac{{n!}}
{{k!\, \cdot \left( {n - k} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!}}
{{(k - 1)!\, \cdot \left( {n - k + 1} \right)!}} \cdot \frac{k}
{k} + \frac{{n!}}
{{k!\, \cdot \left( {n - k} \right)!}} \cdot \frac{{n - k + 1}}
{{n - k + 1}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!\, \cdot k}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} + \frac{{n!\left( {n - k + 1} \right)}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!\, \cdot k + n!\left( {n - k + 1} \right)}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!\left( {k + n - k + 1} \right)}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{n!\left( {n + 1} \right)}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \frac{{\left( {n + 1} \right)!}}
{{k!\, \cdot \left( {n - k + 1} \right)!}} \\
  \left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\

 \end{array} } \right) + \left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right) = \left( {\begin{array}{*{20}c}
   {n + 1}  \\
   k  \\

 \end{array} } \right) \\
\end{gathered}
$