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Formulas Related to Factorials 📂Functions

Formulas Related to Factorials

Product of Consecutive Odd Numbers

For an integer n≥0n \ge 0, the following holds.

(2n−1)⋅(2n−3)⋯5⋅3⋅1=(2n)!2n(n!)=(2n−1)!! (2n-1) \cdot (2n-3) \cdots 5 \cdot 3 \cdot 1 = \dfrac{(2n)!}{2^{n} (n!)} = (2n-1)!!

Here, n!!n!! refers to the double factorial.

Proof

A detailed explanation is omitted.

3⋅1= 4⋅3⋅2⋅14⋅2=4!22(2⋅1)=(2⋅2)!22(2!)5⋅3⋅1= 6⋅5⋅4⋅3⋅2⋅16⋅4⋅2=6!23(3⋅2⋅1)=(2⋅3)!23(3!)7⋅5⋅3⋅1= 8⋅7⋅6⋅5⋅4⋅3⋅2⋅18⋅6⋅4⋅2=8!24(4⋅3⋅2⋅1)=(2⋅4)!24(4!)⋮(2n−1)⋅(2n−3)⋯5⋅3⋅1= (2n)!2n(n!) \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≥0n \ge 0, the following holds.

(2n)⋅(2n−2)⋯6⋅4⋅2=2n(n!) (2n) \cdot (2n-2) \cdots 6 \cdot 4 \cdot 2 = 2^{n}(n!)

Proof

A detailed explanation is omitted.

4⋅2= 22(2⋅1)=22(2!)6⋅4⋅2= 23(3⋅2⋅1)=23(3!)8⋅6⋅4⋅2= 24(4⋅3⋅2⋅1)=24(4!)⋮(2n)⋅(2n−2)⋯6⋅4⋅2= 2n(n!) \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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