Question

Evaluate ${}^{20}{C}_5+{\sum }_{r=2}^5{}^{25-r}{C}_4.$

Answer 1

${}^{20}{C}_5+{\sum }_{r=2}^5{}^{25-r}{C}_4.$

$\Rightarrow ({}^{20}{C}_5+{}^{20}{C}_4)+{}^{21}{C}_4+{}^{22}{C}_4+{}^{23}{C}_4$

$\Rightarrow ({}^{21}{C}_5+{}^{21}{C}_4)+{}^{22}{C}_4+{}^{23}{C}_4$ $(\because {}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r)$
$\Rightarrow ({}^{22}{C}_5+{}^{22}{C}_4)+{}^{23}{C}_4$
$\Rightarrow {}^{23}{C}_5+{}^{23}{C}_4$ $(\because {}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r)$
$\Rightarrow {}^{24}{C}_5$
$\Rightarrow 42504$