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Thread: Help with permutation questions

  1. #1
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    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
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  2. #2
    MHF Contributor
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    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)
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