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

We have, 4n! (2n)!(2n!) 2n! n!n! (4n!) (n!^ 2 ) (2n!)(2n) (2n)!^ 2 [1,2,3,4...(4n-1)(4n) ](n!^ 2 ) (2n)! [1,2,3,4....(2n-2)(2n-1)(2n) ]^ 2 [1,3,5...(4n-1) ] [2,4,6...4n ](n!^ 2 ) (2n)! [1,3,5....(2n-1) ]^ 2 [2,4,6....(2n-2)(2n) ]^ 2 [1,3,5.....(4n-1) ] [1,3,5.....(4n-1) ]^ 2 Hence proved.

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

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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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