7 über 3?
\(\begin{array}{|rcll|} \hline \binom{7}{ 3} &=&\frac{7}{3}\cdot \frac{6}{2}\cdot \frac{5}{1} \\ &=&\frac{7}{3}\cdot 3\cdot \frac{5}{1} \\ &=&7 \cdot \frac{5}{1} \\ &=&7 \cdot 5 \\ &=& 35\\ \hline \end{array}\)
