Question:
If $\sum_{r=0}^{25}\left\{{ }^{50} \mathrm{C}_{\mathrm{r}} \cdot{ }^{50-\mathrm{r}} \mathrm{C}_{25-\mathrm{r}}\right\}=\mathrm{K}\left({ }^{50} \mathrm{C}_{25}\right)$, then $\mathrm{K}$ is equal to:
Correct Option: 4
JEE Main Previous Year Jan. 10, 2019 (II)
Solution: