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I chanced upon this problem:

$\textbf{Show that} \hspace{0.2cm}\frac{(2n)!}{n!} = 2^n(1 \times 3 \times 5 \times ... \times (2n - 1)).$

I tried the following, and realised I was wrong! :

$\frac{(2n)!}{n!} = \frac{2n(2n-1)(2n-2)(2n-3)\times ... \times 1}{n(n-1)(n-2)(n-3)\times ... \times 1}$

$= \frac{2(2n-1)(2n-2)(2n-3)\times ... \times 1}{(n-1)(n-2)(n-3)\times ... \times 1}$

$= 2(2n-1)(2n-3)(2n-5)\times ... \times 1$

$= 2(1 \times 3 t\times 5 \times ... \times (2n-1))$

Where am I going wrong?

Thanks in advance.

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    $\begingroup$ Make sure to realize that $(2n - 2k)/(n-k) = 2$, not $1$. $\endgroup$ – Hyperion Feb 17 at 5:39
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    $\begingroup$ Your approach is fine. Your mistake, as Hyperion mentioned, is when you go from $\frac{2(2n - 1)\ldots 1}{(n - 1)(n - 2)\ldots 1}$ to $2(2n - 1)(2n - 3)(2n - 5)\ldots 1$. Since $(2n - 2k)/(n - k) = 2$, when you cancel out $(2n - 2)/(n - 1)$, or $(2n - 4)/(n - 2)$, a 2 should be left for each of these cancellations. $\endgroup$ – Craveable Banana Feb 17 at 6:14
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    $\begingroup$ @CraveableBanana Thanks a lot. It helped. $\endgroup$ – Ramana Feb 17 at 6:17
  • $\begingroup$ Seems like an obvious candidate to prove by induction. $\endgroup$ – Robert Shore Feb 17 at 8:13
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You can write it so: $$\binom{2n}{n}n!$$ By your way we obtain: $$\frac{(2n)1}{n!}=\frac{2n(2n-1)(2n-2)(2n-3)(2n-4)...1}{n(n-1)(n-2)...1}=$$ $$=2(2n-1)2(2n-3)2(2n-5)...=2^n(2n-1)(2n-3)...1=2^{n}(2n-1)!!$$

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  • $\begingroup$ Could you please elaborate? It would be helpful if you briefly explain the simplification process. $\endgroup$ – Ramana Feb 17 at 5:50
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    $\begingroup$ @Ramana I added something. See now. $\endgroup$ – Michael Rozenberg Feb 17 at 5:54
  • $\begingroup$ So does a double factorial basically mean this: 3!! = 6! = 720? $\endgroup$ – Ramana Feb 17 at 5:55
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Your mistake is that you did not collect the $n$ factors $2$ which would lead to the expression $2^n$ in front.

Writing it as follows may clarify this:

$$\frac{2n!}{n!}=\frac{\prod_{k=1}^{2n}k}{\prod_{k=1}^n k} = \frac{\prod_{i=1}^{n} 2i \cdot \prod_{i=1}^{n}(2i-1) }{\prod_{k=1}^{n}k} = \color{blue}{2^n}\frac{\prod_{i=1}^{n} i \cdot \prod_{i=1}^{n}(2i-1) }{\prod_{k=1}^{n}k}$$ $$= 2^n\prod_{i=1}^{n}(2i-1)$$

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  • $\begingroup$ Thanks. It helped. $\endgroup$ – Ramana Feb 17 at 6:18

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