Question

${\sum }_{r=0}^{25}\left\{{}^{50}{C}_r{\cdot }^{50-r}{C}_{25-r}\}\right\}=K\left({}^{50}{C}_{25}\right)$ then K is equal to:

• A. $(25{)}^2$
• B. $2^{25}-1$
• C. $2^{24}$
• D. ${^{50}\textrm{C}_{25}} $

Answer 1

The correct option is D ${^{50}\textrm{C}_{25}} $

${\sum }_{r=0}^{25}\left\{{}^{50}{C}_r{\cdot }^{50-r}{C}_{25-r}\right\}$

Expanding

${\sum }_{r=0}^{25}\frac{50!}{r!\cdot (50-r)!}\cdot \frac{(50-r)!}{(25-r)!\cdot (50-r-25+r)!}$

${\sum }_{r=0}^{25}\frac{50!}{r!(25-r)!\cdot \left(25\right)!}$

${\sum }_{r=0}^{25}\frac{50!}{r!(25-r)!\cdot \left(25\right)!}\cdot \frac{25!}{25!}$

${\sum }_{r=0}^{25}{}^{50}{\text{C}}_{25}\frac{25!}{r!(25-r)!}$

${\sum }_{r=0}^{25}{{}^{50}{\text{C}}_{25}}^{25}{\text{C}}_r$

${}^{50}{\text{C}}_{25}{\sum }_{r=0}^{25}{}^{25}{\text{C}}_r$

${}^{50}{\text{C}}_{25}\cdot {}^{50}{\text{C}}_{25}$

$\therefore K={}^{50}{\text{C}}_{25}$