Question

 If $\mathop \sum \nolimits_{r = 0}^{25} \left\{ {^{50}{C_r}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)$, then K is equal to-
A. $2^{25} - 1$
B. $25^{2}$
C. $2^{25}$
D. $2^{24}$

Answer 1

Hint:To solve this problem, we need to know the formula of expansion for $^{n}C_{r}$ which is given by-
$\dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ after solving choose the correct answer from options.

Complete step by step answer:
Now, the given expression $^{50}C_{r}$ .$^{50-r}C_{25-r}$ can be re-written as-
$ = \dfrac{{\left( {50} \right)!}}{{\left( {50 - r} \right)!r!}} \times \dfrac{{\left( {50 - r} \right)!}}{{\left( {25} \right)!\left( {25 - r} \right)!}}$
$ = \dfrac{{\left( {50} \right)!}}{{\left( {25} \right)!\left( {25} \right)!}} \times \dfrac{{\left( {25} \right)!}}{{\left( {25 - r} \right)!r!}}$
Now the whole expression can be written as-
$^{50}{C_{25}}.\sum\limits_{r = 0}^{25} {^{25}{C_r}} $
The summation can be expanded as-
$=^{50}C_{25} ({}^{25}C_{0} + {}^{25}C_{1} + {}^{25}C_{2} + … + {}^{25}C_{25})$ …(1)
We know that the binomial expansion $(1 + x)^n = {}^{n}C_{0} + {}^{n}C_{1}x + {}^{n}C_{2}x^2 + … + {}^{n}C_{n}x^n$
At $x = 1$,
$2^n = {}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + … + {}^{n}C_{n}$
At $n = 25$,
$2^{25} = {}^{25}C_{0} + {}^{25}C_{1} + {}^{25}C_{2} + … + {}^{25}C_{25}$ …(2)
So, substituting (2) in (1) the expression can be written as-
$= 2^{25}.{}^{50}C_{25}$
By comparing this expression with K,
$K = 2^{25}$
Hence, the correct option is $C.\text{ } 2^{25}$

Note: In such problems, we have to transform the expression according to what is asked in the question. This comes with practice. For example, in this question, we multiplied and divided by 25! so that we can get the term ${}^{50}C_{25}$.