I am trying to find a simple factorial of all the preceding odd numbers. If $9$ were to be picked the equation would read $9\times 7\times 5\times 3\times 1$ (only odd numbers can be picked). Would the following fraction work?
$$\dfrac{x!}{2^{\left(\frac{x-1}{2}\right)}\left(\frac{x-1}{2}\right)!}$$
On
Hint: \begin{align} 9\cdot7\cdot5\cdot3\cdot1&= \frac{9\cdot8\cdot7\cdot6 \cdot5\cdot4\cdot3\cdot2\cdot1}{8\cdot6\cdot4\cdot2}\\&= \frac{9\cdot8\cdot7\cdot6 \cdot5\cdot4\cdot3\cdot2\cdot1}{2^4(4\cdot3\cdot2\cdot1)}\end{align}
On
Let's think this out. We want:
$N =\frac {(2n + 1)!}{\prod_{m\text{ is even}; m < 2n+1} m}=\frac {(2n + 1)!}{\prod_{k=1}^n 2k}=\frac {(2n + 1)!}{2^n\prod_{k=1}^n k}=\frac {(2n+1)!}{2^n n!}$
Okay, Replace $x=2n + 1$ and $n=\frac {x-1}2$ and we get
$N = \frac {x!}{2^{\frac {x-1}2}(\frac {x-1}2)!}$
But... if you think about this, it doesn't actually make anything easier. There's no reason that $n! = \prod_{k \le n} k$ is any more fundamental a concept or operation than $n!! = \prod_{k\text{ is same even/oddness as }n; k \le n} k$. And we can and do simply define the double factorial as exactly that.
But if you want to convert (for calculation purposes between factorial and double factorials that is the formula:
$n!! =\frac {x!}{2^{\frac {x-1}2}(\frac {x-1}2)!}$ is $x$ is odd and
$n!! =\frac {x!}{2^{\frac {x}2}(\frac {x}2)!}$
or in general $n!! =\frac {x!}{2^{\lfloor \frac {x}2\rfloor}(\lfloor \frac {x}2\rfloor)!}$
where $\lfloor y\rfloor$ is the floor function (greatest integer equal or less than).
That formula doesn't work. To find one that does, notice that $$(2n-1)\times (2n-3) \times \cdots \times 5 \times 3 \times 1$$ is what you get by dividing $(2n)!$ by the following product: $$2n \times (2n-2) \times \cdots \times 6 \times 4 \times 2$$ Each of the terms in this product has a factor of $2$. Pulling all of them out to the front yields a nice closed-form expression in terms of exponents and factorials.