Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& (1919{)}^{1919}=(1920-1{)}^{1919}\\ & ={}^{1919}{\mathrm{C}}_0(1920{)}^{1919}-{}^{1919}{\mathrm{C}}_1(1920{)}^{1918}+\dots .\\ & +{}^{1919}{\mathrm{C}}_{1918}(1920{)}^1-{}^{1919}{\mathrm{C}}_{1919}\\ & =100\lambda +1919\times 1920-1\\ & =100\lambda +3684480-1\\ & =100\lambda +\dots \dots \dots ..79\text{ (last two digit) }\\ & \Rightarrow \text{ Number having last two digit }79\\ & \therefore \text{ Product of last two digit }63\end{array}$