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)

$\displaystyle\sum_{r=0}^{25}\left\{\left({ }^{50} C_{r}\right)\left({ }^{50-r} C_{25-r}\right)\right\}=K\left({ }^{50} C_{25}\right)$
$LHS$
$=\displaystyle\sum_{r=0}^{25} { }^{50} C_{r}{ }^{50-r} C_{25-r}$
$=\displaystyle\sum_{r=0}^{25}\left[\left\{\frac{50 !}{r !(50-r) !}\right\}\left\{\frac{(50-r) !}{(25-r) !(25) !}\right\}\right] $
$=\displaystyle\sum_{r=0}^{25}\frac{50 ! 25 !}{r !(25-r) !(25) !(25) !}$
$={ }^{50} C_{25}\displaystyle\sum_{r=0}^{25}{ }^{25} C_{r}$
$={ }^{50} C_{25} \times 2^{25}$
$=K\left({ }^{50} C_{25}\right)$
$\therefore K=2^{25}$