Question

If $\displaystyle \sum _{r = 0}^{25}\left\{\left(\_{ \, }^{50}C_{r}^{}\right) . \left(\_{ \, }^{50 - r}C_{25 - r}^{}\right)\right\}=K\left(\_{ \, }^{50}C_{25}^{}\right),$ then $K$ is equal to

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

Answer 1

Answer: (A)

LHS $=\displaystyle \sum _{r = 0}^{25} \,{}^{50} C_{r}\,{}^{50 - r}C_{25 - r}$
$=\displaystyle \sum _{r = 0}^{25} \frac{50 !}{r ! \left(50 - r\right) !} \frac{\left(50 - r\right) !}{\left(25 - r\right) ! \left(25\right) !}$
$=\displaystyle \sum _{r = 0}^{25} \, \frac{50 ! \, 25 !}{r ! \left(25 - r\right) ! \left(25\right) ! \left(25\right) !}$
$=\,{}^{50}C_{25} \, \displaystyle \sum _{r = 0}^{25} \,{}^{25} C_{r}$
$=\,{}^{50}C_{25} \, \times 2^{25}$
$\Rightarrow K=2^{25}$