Question

$\sum _{r=1}^{25}{(}^{50}{C}_r{.}^{50-r}{C}_{25-r}\left){=}^{50}{C}_{25}\right(\lambda )$.Then $\lambda$ is

• A. $2^{24}$
• B. $2^{25}$
• C. $2^{25}-1$
• D. $2^{30}$

Answer 1

Answer: (B)

$2^{25}$

${\sum }_{r=0}^{25}{50}_{C_r}{50-r}_{C_{25-r}}=\lambda {50}_{C_{25}}$

LHS :

$={\sum }_{r=0}^{25}\left({50}_{C_r}{50-r}_{C_{25-r}}\right)$

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

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

$={\sum }_{r=0}^{25}(\frac{50!}{25!(25-r)!}\times \frac{1}{r!}\times \frac{25!}{25!})$

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

$={50}_{C_{25}}\left({\sum }_{r=0}^{25}{25}_{C_r}\right)$

Now, ${\sum }_{r=0}^{25}{25}_{C_r}$ is basically the sum of binomial coefficients of the expansion ${(1+1)}^{25}=2^{25}$

$={50}_{C_{25}}.2^{25}$

Comparing with RHS we get $\lambda =2^{25}$

Answer : Option B.