Question

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

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

Answer 1

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

$\sum _{r=0}^{25}$ ${}^{50}{C}_r{\cdot }^{50-r}{C}_{25-r}$

$=\sum _{r=0}^{25}\frac{50!}{r!(50-r)!}\times \frac{(50-r)!}{\left(25\right)!(25-2)!}$

$=\sum _{r=0}^{25}\frac{50!}{25!25!}\times \frac{25!}{(25-r)!(r!)}$

$={}^{50}{C}_{25}\sum _{r=0}^{25}$ ${}^{25}{C}_r=\left(2^{25}\right){}^{50}{C}_{25}$

$\therefore K=2^{25}$