Question

The value of $\sum _{r=0}^{10}(\frac{10}{r})(\frac{15}{14-r})$ is equal to

• A. ${}^{25}{C}_{12}$
• B. ${}^{25}{C}_{15}$
• C. ${}^{25}{C}_{10}$
• D. ${}^{25}{C}_{11}$

Answer 1

The correct option is D ${}^{25}{C}_{11}$

We know that
${}^n{C}_0{}^m{C}_r+{}^n{C}_1{}^m{C}_{r-1}+{}^n{C}_2{}^m{C}_{r-2}+...{}^n{C}_r{}^m{C}_0$
$={}^{n+m}{C}_r$
Applying to the above, we get
$S_n$
$={}^{10+15}{C}_{14-r+r}$
$={}^{25}{C}_{14}$
$={}^{25}{C}_{11}$.