# Thread: Help with permutation questions

1. ## Help with permutation questions

a) Prove that $\displaystyle \frac{(2n)!}{n!} = 2^n . {1.3.5.7.....(2n-1)}$

b) Prove that $\displaystyle 2.6.10.14......(4n-6)(4n-2)$ equal $\displaystyle (n+1)(n+2)(n+3).....(2n-1)2n$

for part a i have
=$\displaystyle \frac{(2n)(2n-1)(2n-3)......n(n-1)(n-2)(n-3)........}{n(n-1)(n-2)(n-3).....}$
= $\displaystyle (2n)(2n-1)(2n-3).....$

how do i get from $\displaystyle (2n)(2n-1)(2n-3).....$ to
$\displaystyle 2^n . {1.3.5.7.....(2n-1)}$

Please can some tell me how to get to the answer. I am finding this very difficult

for part b i do not know what to do

2. ## Re: Help with permutation questions

use induction

for (a)

$\dfrac{2!}{1!} = 2^1 \cdot 1$

so $P_1=True$

assume $P_n$ is true

\begin{align*} &\dfrac{(2(n+1))!}{(n+1)!} = \\ \\ &\dfrac{(2n+2)(2n+1)(2n)!}{(n+1)n!} = \\ \\ &\dfrac{(2n+2)(2n+1)}{n+1}\cdot \dfrac{(2n)!}{n!} = \\ \\ &[2(2n+1)]\cdot 2^n \prod\limits_{k=0}^{n-1} (2k+1) = \\ \\ &2^{n+1} \prod\limits_{k=0}^{n} (2k+1) \end{align*}

and thus $P_{n+1}=True$

so by induction $P_n = True,~\forall n \in \mathbb{N}$

Apply this same method to (b)