JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 14)
If $$\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$$ then K is equal to :
224
225$$-$$ 1
225
(25)2
Explanation
$$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$
$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$
$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$
$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$
$$ \therefore $$ $$K = {2^{25}}$$
$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$
$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$
$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$
$$ \therefore $$ $$K = {2^{25}}$$
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