Double Factorials and Multifactorials

Before reviewing this article, the readers are expected to know what factorials are first.

The double factorial of a positive integer $n$ is the generalization of the factorial $n!;$ 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, and $1$ 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 $n=0$, then $0!!=1.$

To clarify, $n!!$ is not equal to $(n!)!$. For example, $4!! = 4\times2=8,$ whereas $(4!)! = 24! = 620448401733239439360000$. Go ahead and try out the following warm-up problem:

For any non-negative integer $n,$ we find that

$$\dfrac{n!}{n!!}=(n-1)!! ~~\text{ or }~~ n!=(n-1)!!×n!!.$$

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 any non-negative integer $n$, $$\dfrac{n!}{n!!}=(n-1)!!. \ _\square$$

For any non-negative integer $n,$ we find that

$$\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!.$$

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 $\frac {9!}{9!!}$.

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Since $\frac {n!}{n!!}=(n-1)!!$, substituting the values, we get

$$\begin{aligned} \dfrac{9!}{9!!}&=(9-1)!!\\ &=8!!\\ &=8×6×4×2\\ &=384. \ _\square \end{aligned}$$

Evaluate $\frac {(3!)!}{3!!}$.

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We have

$$\begin{aligned} \dfrac {(3!)!}{3!!} &=\dfrac {(3×2×1)!}{3×1}\\ &=\dfrac {6!}{3}\\ &=\dfrac {6×5×4×3×2×1}{3}\\ &=\dfrac {720}{3}\\ &=240. \ _\square \end{aligned}$$

For any non-negative integer $n$ we find that

$$\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!.$$

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 $\frac{9!}{8!!}$.

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We have

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