Question

$\text { If } \sum_{r=0}^{25}\left\{{ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right\}=K\left({ }^{50} C_{25}\right) \text {, then } K \text { is equal to : }$

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

Answer 1

Answer: (C)

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