Question

The sum of the series ${\sum }_{i=1}^{50}{{}^{100}C}_{50-i}{{}^{50}C}_i$ equals

• A. ${{}^{150}C}_{50}$
• B. ${\left({{}^{100}C}_{50}\right)}^2$
• C. ${{}^{150}C}_{50}-{{}^{100}C}_{50}$
• D. None of these

Answer 1

The correct option is A ${{}^{150}C}_{50}$

The Coefficient of $x^k$ in ${(1+x)}^{m+n}$ is same as the coefficient in ${(1+x)}^m{(1+x)}^n$,

$⟹\left(\begin{array}{c}m+n\\ k\end{array}\right)={\sum }_{r=0}^k\left(\begin{array}{c}m\\ k\end{array}\right)\left(\begin{array}{c}n\\ k-r\end{array}\right)$

${\sum }_{i=1}^{50}\left(\begin{array}{c}100\\ 50-i\end{array}\right)\left(\begin{array}{c}50\\ i\end{array}\right)=\left(\begin{array}{c}150\\ 50\end{array}\right)$

The above theorem is also called $VANDERMOND{E}^{\prime }STHEOREM$