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
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.