Formulas Related to Factorials
Product of Consecutive Odd Numbers
For an integer $n \ge 0$, the following holds.
$$ (2n-1) \cdot (2n-3) \cdots 5 \cdot 3 \cdot 1 = \dfrac{(2n)!}{2^{n} (n!)} = (2n-1)!! $$
Here, $n!!$ refers to the double factorial.
Proof
A detailed explanation is omitted.
$$ \begin{align*} 3 \cdot 1 =&\ \dfrac{4 \cdot 3 \cdot 2 \cdot 1}{4 \cdot 2} = \dfrac{4!}{2^{2}(2 \cdot 1)} = \dfrac{(2 \cdot 2)!}{2^{2}(2!)} \\ 5 \cdot 3 \cdot 1 =&\ \dfrac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 4 \cdot 2} = \dfrac{6!}{2^{3}(3 \cdot 2 \cdot 1)} = \dfrac{(2 \cdot 3)!}{2^{3}(3!)} \\ 7 \cdot 5 \cdot 3 \cdot 1 =&\ \dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{8 \cdot 6 \cdot 4 \cdot 2} = \dfrac{8!}{2^{4}(4 \cdot 3 \cdot 2 \cdot 1)} = \dfrac{(2 \cdot 4)!}{2^{4}(4!)} \\ \vdots& \\ (2n-1) \cdot (2n-3) \cdots 5 \cdot 3 \cdot 1 =&\ \dfrac{(2n)!}{2^{n}(n!)} \end{align*} $$
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Product of Consecutive Even Numbers
For an integer $n \ge 0$, the following holds.
$$ (2n) \cdot (2n-2) \cdots 6 \cdot 4 \cdot 2 = 2^{n}(n!) $$
Proof
A detailed explanation is omitted.
$$ \begin{align*} 4 \cdot 2 =&\ 2^{2}(2 \cdot 1) = 2^{2}(2!) \\ 6 \cdot 4 \cdot 2 =&\ 2^{3}(3 \cdot 2 \cdot 1) = 2^{3}(3!) \\ 8 \cdot 6 \cdot 4 \cdot 2 =&\ 2^{4}(4 \cdot 3 \cdot 2 \cdot 1) = 2^{4}(4!) \\ \vdots& \\ (2n) \cdot (2n-2) \cdots 6 \cdot 4 \cdot 2 =&\ 2^{n}(n!) \end{align*} $$
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