MathematicsMedium98×since 2002Q4823If ∑r=025{50Cr.50−rC25−r}=K(50C25), \sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,r=0∑25{50Cr.50−rC25−r}=K(50C25), then K is equal to :A2²⁴B2²⁵−-− 1C2²⁵D(25)²Check answerSkip