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Mathematics Question on Binomial theorem

If \sum^\limits{25}_{r=0} \left\\{^{50}C_{r} . ^{50-r}C_{25-r}\right\\}=K\left(^{50}C_{25}\right) , then KK is equal to :

A

225−12^{25} - 1

B

(25)2(25)^2

C

2252^{25}

D

2242^{24}

Answer

2252^{25}

Explanation

Solution

∑r=02550Cr.50−rC25−r\sum^{25}_{r=0} {^{50}C_{r}} . {^{50-r}C_{25-r}}
=∑r=02550!r!(50−r)!×(50−r)!(25)!(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)!}
=∑r=02550!25!25!×25!(25−r)!(r!)=\sum^{25}_{r=0} \frac{50!}{25!25!} \times\frac{25!}{\left(25-r\right)!\left(r!\right)}
=50C25∑r=02525Cr=(225)50C25= {^{50}C_{25}} \sum^{25}_{r=0} {^{25}C_{r}} =\left(2^{25}\right)^{50}C_{25}
∴K=225\therefore K = 2^{25}