Factorial equality $\ \frac{1\cdot3\cdot5\cdots(2k-1)}{2^k2!}$ $\ = \frac{(2k)!}{2^k2^kk!k!}$

Question

In a generating function identity proof in my textbook there is a step that I can't wrap my head around.

$$ \frac{1\cdot3\cdot5\cdots(2k-1)}{2^k2!}$$ $$ = \frac{(2k)!}{2^k 2^k k!k!}$$

How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?

Answer 1

$$\prod_{r=1}^k(2r-1)=\dfrac{\prod_{r=1}^{2k}r}{2^k\cdot k!}=?$$

Answer 2

HINT:

Note that we can write

$$1\cdot3\cdot 5\cdots (2k-1)=\frac{1\cdot 2\cdot 3\cdot 4\cdots (2k-2)(2k-1)(2k)}{2(1)\cdot 2(2)\cdots 2(k-1)2(k)}=\frac{(2k)!}{2^k\,k!}$$

Answer 3

\begin{align} \frac{1\cdot3\cdot5\cdot7\cdot11\cdot13}{2^7\cdot2!} & = \frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14}{2^7 \cdot2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14} \\[10pt] & = \frac{14!}{2^7\cdot(2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2)\cdot(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7)} = \frac{14!}{2^7\cdot2^7\cdot7!} \end{align}

Answer 4

In the denominator of the lhs, $2!$ should be $k!$

Let $1\cdot 3\cdot 5\cdot \cdot \cdot (2k-1) = (2k-1)!!\,$, the double factorial.

We want to show that \begin{equation} (2k-1)!! = \frac{(2k)!}{2^{k}k!} \end{equation}

We have \begin{align} (2k-1)!!2^{k}k! &= [(2k-1)(2k-3)\cdot \cdot \cdot 1]2^{k}k(k-1)(k-2)\cdot \cdot \cdot 1 \\ \tag{1} &= [(2k-1)(2k-3)\cdot \cdot \cdot 1][2k][2(k-1)][2(k-2)]\cdot \cdot \cdot [2\cdot 1] \\ \tag{2} &= 2k(2k-1)(2k-2)\cdot \cdot \cdot 1\\ &= (2k)! \end{align}

1. Associate each $2$ from $2^{k}$ with each term of $k!$

2. Rearrange terms.