Question

If $\sum _{r=0}^{25}\{{}^{50}{C}_r\cdot {}^{50-r}{C}_{25-r}\}=K\left({}^{50}{C}_{25}\right)$ then $K$ is equal to :

• A. ${25}^2$
• B. $2^{24}$
• C. $2^{25}-1$
• D. $2^{25}$

Answer 1

The correct option is D $2^{25}$

$\sum _{r=0}^{25}{}^{50}{C}_r{}^{50-r}{C}_{25-r}$
$=\sum _{r=0}^{25}\frac{50!}{r!(50-r)!}\times \frac{(50-r)!}{25!(25-r)!}$
$=\sum _{r=0}^{25}\frac{50!25!}{r!(25-r)!25!25!}$
$={}^{50}{C}_{25}\sum _{r=0}^{25}{}^{25}{C}_r$
$={}^{50}{C}_{25}\times 2^{25}$
Hence, $K=2^{25}$