Question

Prove that $\dfrac{\left( 2n \right)!}{n!}={{2}^{n}}\left( 1\times 3\times 5\times \cdots \left( 2n-1 \right) \right)$

Answer 1

Hint: Use the fact that $n!=n\left( n-1 \right)!$. Hence write (2n)! as (2n)(2n-1)(2n-2)…(2)(1)
Now from every even term(2n,2n-2,2n-4,…) take 2 as common. Argue that there will be n such terms. Hence prove that $\left( 2n \right)!={{2}^{n}}\left( 1\times 3\times 5\times \cdots \left( 2n-1 \right) \right)\left( 1\times 2\times \cdots \times n \right)$
Divide both sides by n! and hence prove that $\dfrac{\left( 2n \right)!}{n!}={{2}^{n}}\left( 1\times 3\times 5\times \cdots \times \left( 2n-1 \right) \right)$

Complete step-by-step answer:
We know that $n!=n\left( n-1 \right)!$
Hence, we have $\left( 2n \right)!=2n\left( 2n-1 \right)!$
Continuing this way, we get
$\left( 2n \right)!=2n\left( 2n-1 \right)\left( 2n-2 \right)\cdots \left( 2 \right)\left( 1 \right)$
Writing even terms first and then the odd terms(This can be done since multiplication is both commutative and associative), we get
$\left( 2n \right)!=\left[ \left( 2n \right)\left( 2n-2 \right)\left( 2n-4 \right)\cdots 2 \right]\left[ \left( 2n-1 \right)\left( 2n-3 \right)\cdots \left( 1 \right) \right]$
Taking 2 common from each of the even terms, we get
$\left( 2n \right)!=\left[ 2\left( n \right)\left( 2 \right)\left( n-1 \right)\left( 2 \right)\left( n-2 \right)\cdots \left( 2 \right)\left( 1 \right) \right]\left[ \left( 2n-1 \right)\left( 2n-3 \right)\cdots \left( 1 \right) \right]$
From 1 to n, we have n terms. Hence there are n 2s in the product.
Hence, we have
$\left( 2n \right)!=\left[ {{2}^{n}}\left( 1\times 2\times \cdots \times n \right) \right]\left[ \left( 2n-1 \right)\left( 2n-3 \right)\cdots \left( 1 \right) \right]$
We know that $1\times 2\times 3\cdots \left( n \right)=n!$
Hence, we have
$\left( 2n \right)!=\left[ {{2}^{n}}n! \right]\left[ \left( 2n-1 \right)\left( 2n-3 \right)\cdots \left( 1 \right) \right]$
Dividing both sides by n!, we get
$\dfrac{\left( 2n \right)!}{n!}={{2}^{n}}\left( 1\times 3\times 5\times \cdots \left( 2n-1 \right) \right)$
Q.E.D

Note: Alternative Solution:
We know that the power of prime p in n! is given by $e=\left[ \dfrac{n}{p} \right]+\left[ \dfrac{n}{{{p}^{2}}} \right]+\cdots $, where [x] denotes the greatest integer less or equal to x.
Since 2 is prime, we have
Power of 2 in (2n)! is equal to $\left[ \dfrac{2n}{2} \right]+\left[ \dfrac{2n}{{{2}^{2}}} \right]+\left[ \dfrac{2n}{{{2}^{3}}} \right]+\cdots $
Similarly, we have the power of 2 in n! is equal to $\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{{{2}^{2}}} \right]+\cdots $
Hence, the power of 2 in $\dfrac{\left( 2n \right)!}{n!}$ is equal to $\left[ \dfrac{2n}{2} \right]+\left[ \dfrac{2n}{{{2}^{2}}} \right]+\left[ \dfrac{2n}{{{2}^{3}}} \right]+\cdots -\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{{{2}^{2}}} \right]+\cdots $
Hence, we have
Power of 2 in $\dfrac{\left( 2n \right)!}{n!}$ is equal to
$n+\left[ \dfrac{n}{2} \right]+\left[ \dfrac{n}{{{2}^{2}}} \right]+\cdots -\left[ \dfrac{n}{2} \right]-\left[ \dfrac{n}{{{2}^{2}}} \right]\cdots =n$
The remaining terms in $\dfrac{\left( 2n \right)!}{n!}$ are all odd numbers from 1 to 2n-1.
Hence, we have
$\dfrac{\left( 2n \right)!}{n!}={{2}^{n}}\left( 1\times 3\times 5\times \cdots \left( 2n-1 \right) \right)$
Hence proved.