Question Number 154514 by 0731619 last updated on 19/Sep/21
Commented by mr W last updated on 19/Sep/21
$${there}\:{is}\:{not}\:{much}\:{to}\:{solve}!\:{it}'{s}\:{a}\: \\ $$$${matter}\:{of}\:{definition}. \\ $$$$\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$