Prove that $\dfrac{(2n)!}{n!} = 2^n(1.3.5\ldots(2n-1))$

Prove that ((2n)!)/(n!) = 2ⁿ(1.3.5…(2n-1))

Question 134 of 134

\dfrac{(2n)!}{n!} = 2^n(1.3.5\ldots(2n-1))

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2026Subjective· 3mImportance★★★★★

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Writing $(2n)!$ as the product of its odd factors times its even factors, and pulling a 2 out of each even factor, directly yields $2^n\cdot n!\cdot(1\cdot3\cdots(2n-1))$ , which rearranges to the required identity.

$(2n)!=1\cdot2\cdot3\cdot4\cdot5\cdots(2n-1)(2n)$

Separate the odd-positioned and even-positioned factors:

$(2n)!=\underbrace{[1\cdot3\cdot5\cdots(2n-1)]}_{\text{odd factors}}\times\underbrace{[2\cdot4\cdot6\cdots(2n)]}_{\text{even factors}}$

The even factors can be written as $2\cdot4\cdot6\cdots(2n)=2^n(1\cdot2\cdot3\cdots n)=2^n\cdot n!$ (factoring a 2 out of each of the $n$ even terms).

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Q.Prove that $\dfrac{(2n)!}{n!} = 2^n(1.3.5\ldots(2n-1))$

Concept understanding — Factorials

Q.Prove that (2n)!n!=2n(1.3.5…(2n−1))\dfrac{(2n)!}{n!} = 2^n(1.3.5\ldots(2n-1))n!(2n)!​=2n(1.3.5…(2n−1))

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Q.Prove that $\dfrac{(2n)!}{n!} = 2^n(1.3.5\ldots(2n-1))$

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