Question:
If $\sum_{\mathrm{r}=0}^{25}\left\{{ }^{50} \mathrm{C}_{\mathrm{r}} \cdot{ }^{50-\mathrm{r}} \mathrm{C}_{25-\mathrm{r}}\right\}=\mathrm{K}\left({ }^{50} \mathrm{C}_{25}\right)$, then $\mathrm{K}$ is equal to :
Correct Option: , 3
Solution:
$\sum_{r=0}^{25}{ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}$
$=\sum_{r=0}^{25} \frac{50 !}{r !(50-r) !} \times \frac{(50-r) !}{(25) !(25-r) !}$
$=\sum_{r=0}^{25} \frac{50 !}{25 ! 25 !} \times \frac{25 !}{(25-r) !(r !)}$
$={ }^{50} \mathrm{C}_{25} \sum_{\mathrm{r}=0}^{25}{ }^{25} \mathrm{C}_{\mathrm{r}}=\left(2^{25}\right){ }^{50} \mathrm{C}_{25}$
$\therefore \mathrm{K}=2^{25}$
Option (3)