Question:
If $\sum_{r=0}^{25}\left\{{ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right\}=K\left({ }^{50} C_{25}\right)$, then $K$ is equal to:
Correct Option: , 4
Solution:
$\sum_{r=0}^{25}\left({ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right)=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{|50-r| r} \frac{\mid 50-r}{|25| 25-r}\right)$
$=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{\lfloor 25} \times \frac{1}{\lfloor 25} \times\left(\frac{\lfloor 25}{\lfloor 25-r \mid r}\right)\right)$