${\sum _{r=0}^{10}}^{(40-r)}{C}_5=$

$$\begin{gathered} \sum _{\mathrm{r}=0}^{10} {}^{(40-\mathrm{r})}\mathrm{C}_5 \\ =40{\mathrm{C}}_5+39{\mathrm{C}}_5+38{\mathrm{C}}_5+ ... + 30{\mathrm{C}}_5 \\ =\left(30{\mathrm{C}}_6+30{\mathrm{C}}_5\right)+31{\mathrm{C}}_5+32{\mathrm{C}}_5+ ... + 40{\mathrm{C}}_5-30{\mathrm{C}}_6 \\ =\left(31{\mathrm{C}}_6+31{\mathrm{C}}_5\right)+32{\mathrm{C}}_5+ ... + 40{\mathrm{C}}_5-30{\mathrm{C}}_6 \\ =\left(32{\mathrm{C}}_6+32{\mathrm{C}}_5\right)+ ... + 40{\mathrm{C}}_5-30{\mathrm{C}}_6 \\ =40{\mathrm{C}}_6+40{\mathrm{C}}_5-30{\mathrm{C}}_6 \\ ={41}_{{\mathrm{C}}_6}-{30}_{{\mathrm{C}}_6} \end{gathered}$$