Question:
If $\sum^\limits{25}_{r=0} \left\{^{50}C_{r} . ^{50-r}C_{25-r}\right\}=K\left(^{50}C_{25}\right) $ , then $K$ is equal to :
If $\sum^\limits{25}_{r=0} \left\{^{50}C_{r} . ^{50-r}C_{25-r}\right\}=K\left(^{50}C_{25}\right) $ , then $K$ is equal to :
Updated On: Feb 14, 2025
- $2^{25} - 1$
- $(25)^2$
- $2^{25}$
- $2^{24}$
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The Correct Option is C
Solution and Explanation
$\sum^{25}_{r=0} {^{50}C_{r}} . {^{50-r}C_{25-r}} $
$ =\sum^{25}_{r=0} \frac{50!}{r!\left(50-r\right)!} \times\frac{\left(50-r\right)!}{\left(25\right)!\left(25-r\right)!} $
$ =\sum^{25}_{r=0} \frac{50!}{25!25!} \times\frac{25!}{\left(25-r\right)!\left(r!\right)} $
$ = {^{50}C_{25}} \sum^{25}_{r=0} {^{25}C_{r}} =\left(2^{25}\right)^{50}C_{25} $
$ \therefore K = 2^{25} $
$ =\sum^{25}_{r=0} \frac{50!}{r!\left(50-r\right)!} \times\frac{\left(50-r\right)!}{\left(25\right)!\left(25-r\right)!} $
$ =\sum^{25}_{r=0} \frac{50!}{25!25!} \times\frac{25!}{\left(25-r\right)!\left(r!\right)} $
$ = {^{50}C_{25}} \sum^{25}_{r=0} {^{25}C_{r}} =\left(2^{25}\right)^{50}C_{25} $
$ \therefore K = 2^{25} $
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
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