If _r=0^25^50C_r^50-rC_25-r=K(^50C_25), then K is equal to:

Question

If $\displaystyle\sum_{r=0}^{25}\left\{{}^{50}C_r\cdot{}^{50-r}C_{25-r}\right\}=K\left({}^{50}C_{25}\right)$, then $K$ is equal to:

Accepted Answer

$$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}}  \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$

$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$

$$ \therefore $$  $K = {2^{25}}$

Solutions · 1