# Double Factorials and multifactorials

#### Contents

## Double Factorial

This type of factorial is denoted by \(n!!\). It is a type of multifactorial which will be discussed later. As far as double factorial is concerned, it ends with \(2\) for an even number, ends with \(1\) for an odd number:

For an

**even**number & \(n>0\), \(n!!=n\times (n-2)\times \cdots\times 4\times 2.\)For an

**odd**number & \(n>0\), \(n!!=n\times (n-2)\times \cdots\times 3\times 1.\)If \(n=0\), \(0!!=1.\)

## Some useful theorems

For any non-negative integer \(n,\) we find that

\[\dfrac{n!}{n!!}=(n-1)!! ~~\text{ or }~~ n!=(n-1)!!×n!!. \ _\square\]

We have the following 2 cases:

- If \(n\) is
odd, \[\dfrac{n!}{n!!}=\dfrac{n\times (n-1)\times (n-2)\times \cdots \times 3\times 2\times 1}{n\times (n-2)\times (n-4)\times \cdots 5\times 3\times 1}.\] Since all the odd numbers \(n, n-2, n-4, \ldots , 5, 3\) get canceled, we are left with the equation \[\dfrac{n!}{n!!}=(n-1)!!.\]- If \(n\) is
even, \[\dfrac{n!}{n!!}=\dfrac{n\times (n-1)\times (n-2)\times \cdots \times 3\times 2\times 1}{n\times (n-2)\times (n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(n, n-2, n-4, \ldots , 4, 2\) get canceled, we are left with the equation \[\dfrac{n!}{n!!}=(n-1)!!.\]Combining both cases, we find that for

anynon-negative integer \(n\), \[\dfrac{n!}{n!!}=(n-1)!!. \ _\square\]

Suppose that \(n!!\) is defined as follows:

\[ n!! = \begin{cases} n \times (n-2) \times \cdots \times 5 \times 3 \times 1 &\text{if } n \text{ is odd.} \\ n \times (n-2) \times \cdots \times 6 \times 4 \times 2 &\text{if } n \text{ is even.} \\ 1 &\text{if } n = 0, - 1. \\ \end{cases} \]

Then what is

\[\color{red}{\dfrac{9!}{6!!}} \div \color{green}{\dfrac{9!!}{6!}}?\]

For any non-negative integer \(n,\) we find that:\[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!. \ _\square\]

Here, there is no need to consider two separate cases because it makes no difference whether \(n\) is odd or even.

We can expand the LHS as \[\dfrac{(2n+1)\times (2n)\times (2n-1)\times \cdots \times 3\times 2\times 1}{(2n)\times (2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(2n, 2n-2, 2n-4, \ldots, 4, 2\) get canceled, we are left with the equation \[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!. \ _\square\]

Evaluate \(\dfrac {9!}{9!!}\).

Since \(\dfrac {n!}{n!!}=(n-1)!!\), substituting the values, we get \[\begin{align} \dfrac{9!}{9!!}&=(9-1)!!\\ &=8!!\\ &=8×6×4×2\\ &=384. \ _\square \end{align}\]

Evaluate \(\dfrac {(3!)!}{3!!}\).

We have \[\begin{align} \dfrac {(3!)!}{3!!} &=\dfrac {(3×2×1)!}{3×1}\\ &=\dfrac {6!}{3}\\ &=\dfrac {6×5×4×3×2×1}{3}\\ &=\dfrac {720}{3}\\ &=240. \ _\square \end{align}\]

\[\Large{\color{green}{\dfrac{9!}{8!!}}} \div {\color{orange}{\dfrac{7!}{6!!}}} = \ ? \]

**Notation:**

\[ n!! = \begin{cases} n \times (n-2) \times \cdots \times 5 \times 3 \times 1 , && \text{if } n \text{ is odd.} \\
n \times (n-2) \times \cdots \times 6 \times 4 \times 2 , && \text{if } n \text{ is even.} \\
1, && \text{if } n = 0, - 1. \\ \end{cases} \]

\[\]

##### Try the first part here!

For any non-negative integer \(n\) we find that

\[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!. \ _\square\]

Again here, there is no need to consider two separate cases. We can expand the LHS as \[\dfrac{(2n-1)\times (2n-2)\times (2n-3)\times \cdots \times 3\times 2\times 1}{(2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(2n-2, 2n-4, \ldots , 4, 2\) get canceled, we are left with the equation \[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!. \ _\square\]

Evaluate \(\dfrac{9!}{8!!}\).

We have \[ \begin{align}

\dfrac { 9! } { 8!!} &= \dfrac{ (2 \times 5 - 1 ) ! } { (2 \times 5 - 2 ) !! } \\ &= (2 \times 5 - 1 )!! \\ &= 9 !! \\ &= 9 \times 7 \times 5 \times 3 \times 1 \\ &= 945. \ _\square \end{align} \]

**Note:**Do not interpret \(n!!\) to be \((n!)!\). They are completely different!

**Cite as:**Double Factorials and multifactorials.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/double-factorials-and-multifactorials/