Equality of binomial coefficients I have seen that the following equations are equal, but are wondering how this is shown
${n \choose m} \cdot 1 \cdot 3 \cdots (2m-1)\cdot 1 \cdot 3 \cdots (2n-2m-1)
= \frac{n!}{2^n} {2m \choose m} {2n-2m \choose n-m} 
=2^n \cdot n! \cdot (-1)^n \cdot {-\frac{1}{2} \choose m}{-\frac{1}{2} \choose n-m}$
I can show that 
${n \choose m} \cdot 1 \cdot 3 \cdots (2m-1)\cdot 1 \cdot 3 \cdots (2n-2m-1) =2^n \cdot n! \cdot (-1)^n \cdot {-\frac{1}{2} \choose m}{-\frac{1}{2} \choose n-m} $
But I cannot show one of the two equalities involving $\frac{n!}{2^n} {2m \choose m} {2n-2m \choose n-m}$.
I know it is true due to numerical calculations. 
Thanks,

Cool question. I think the following trick works:
$$ {2m \choose m} = \frac{1 \cdot 3 \cdots (2m-1)}{m!} \cdot \frac{2 \cdot 4 \cdots 2m}{1 \cdot 2 \cdots m} = 1 \cdot 3 \cdots (2m-1) \cdot \frac{2^m}{m!} $$
and
$$ {2n-2m \choose n-m} = \frac{1 \cdot 3 \cdots (2n - 2m -1)}{(n-m)!} \cdot \frac{2 \cdot 4 \cdots (2n-2m)}{1 \cdot 2 \cdots (n-m)} = 1 \cdot 3 \cdots (2n - 2m -1) \cdot \frac{2^{n-m}}{(n-m)!}$$
So 
$$ \frac{n!}{2^n} {2m \choose m}{2n-2m \choose n-m} = 1 \cdot 3 \cdots (2m-1) \cdot 1 \cdot 3 \cdots (2n - 2m -1) \cdot \frac{2^{m+n-m}}{2^n} \cdot \frac{n!}{m! (n-m)!} $$
$$ = {n \choose m} \cdot 1 \cdot 3 \cdots (2m-1) \cdot 1 \cdot 3 \cdots (2n - 2m -1)   $$
