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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 :

JEE · Math · previous-year question

  1. A.224
  2. B.225$$-$$ 1
  3. C.225correct
  4. D.(25)2

Answer

C. 225

Explanation

$$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$ $$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$ $$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$ $$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$ $$ \therefore $$ $$K = {2^{25}}$$

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