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This was in one of the examples of the textbook, but I couldn't figure out how they solved it. They say they multiply the left hand side by $\frac{n!}{n!}$ to get the right hand side:

$$ \frac{2^n \cdot (2n-3)!!}{n!} = 2\frac{(2n-2)!}{n!(n-1)!} $$

The double factorial stands for the product of all odd integers from 1 to 2$n$-3.

I've given this problem a lot of time, and I'm all out of ideas at this point.

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@Clive: I think this is fairly standard notation, and it is used in the common references en.wikipedia.org/wiki/Factorial#Double_factorial and mathworld.wolfram.com/DoubleFactorial.html (I don't know if there is another common notation for it.) –  Jonas Meyer Jan 2 '13 at 22:48
    
@JonasMeyer: Oh wow, I stand corrected! Thanks for pointing that out, it'd have been even more embarrassing in a few years :) –  Clive Newstead Jan 2 '13 at 22:51
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@CliveNewstead Though it is a fairly standard notation, I believe it is a very poor notation. So there is no reason for you to feel embarrassed by this poor notation :). –  user17762 Jan 2 '13 at 22:52

2 Answers 2

up vote 6 down vote accepted

Notice that $$2^n n! = 2(n) \cdot 2(n-1) \cdot 2(n-2) \cdots 2(2) \cdot 2(1) = (2n)(2n-2) \cdots (4)(2)$$

is the product of all even integers between $1$ and $2n$. So when you multiply the numerator and denominator by $n!$ you end up with

$$\dfrac{2n(2n-2)!}{n!n!}$$

since the even terms between $1$ and $2n-2$ interweave between the odd terms.

Cancelling the rogue $n$ on the numerator gives the result you seek.

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All you need is: $$ K:=2^{n-1}(n-1)!=(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdots (2\cdot(n-1))=2\cdot 4\cdots (2n-2). $$ This gives for $K$ the product of all even integers between $1$ and $2n-2$. Therefore: $$(2n-2)!/K=(2n-3)!!$$ And this why your LHS turns into your RHS.

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