Prove that: ^ 4n C_ 2n :^ 2n C_ n =[1,3,5...(4n-1):[1,3,5...(2n-1… - Vidyadip

Question

Prove that: $^{4\text{n}}\text{C}_{2\text{n}}:^{2\text{n}}\text{C}_{\text{n}}=[1,3,5...(4\text{n}-1):[1,3,5...(2\text{n-1})^{2}].$

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Answer

We have,
$\Rightarrow \frac{\frac{4\text{n}!}{(2\text{n})!(2\text{n}!)}}{\frac{2\text{n}!}{\text{n}!\text{n}!}}$
$\Rightarrow \frac{(4\text{n}!)\times(\text{n}!^{2})}{(2\text{n}!)(2\text{n})\times(2\text{n})!^{2}}$
$\Rightarrow \frac{\big[1,2,3,4...(4\text{n}-1)(4\text{n})\big](\text{n}!^{2})}{(2\text{n})!\big[1,2,3,4....(2\text{n}-2)(2\text{n}-1)(2\text{n})\big]^{2}}$
$\Rightarrow \frac{\big[1,3,5...(4\text{n}-1)\big]\times\big[2,4,6...4\text{n}\big](\text{n}!^{2})}{(2\text{n})!\big[1,3,5....(2\text{n}-1)\big]^{2}\times\big[2,4,6....(2\text{n}-2)(2\text{n})\big]^{2}}$
$\Rightarrow \frac{\big[1,3,5.....(4\text{n}-1)\big]}{\big[1,3,5.....(4\text{n}-1)\big]^{2}}$
Hence proved.

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