• # question_answer 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: A) ${{(25)}^{2}}$ B) ${{2}^{25}}-1$ C) ${{2}^{24}}$ D) ${{2}^{25}}$

Correct Answer: B

Solution :

 $\sum\limits_{r=\,\,1}^{25}{\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 50-r \,}} \right. }}\times \frac{\left| \!{\underline {\, 50-r \,}} \right. }{\left| \!{\underline {\, 25-r \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }$ $=\sum\limits_{r=\,\,1}^{25}{\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 50-r \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }}=\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, 25 \,}} \right. }\sum\limits_{r=\,\,1}^{25}{\frac{1}{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 25-r \,}} \right. }}$ $=\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, 25 \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }\sum\limits_{r=\,\,1}^{25}{{}^{25}{{C}_{r}}={}^{50}{{C}_{25}}}({{2}^{25}}-1).$

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