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×(n−2)×⋯×4×2;
for an odd number n>0, n!!=n×(n−2)×⋯×3×1;
if n=0, then 0!!=1.
To clarify, n!! is not equal to (n!)!. For example, 4!!=4×2=8, whereas (4!)!=24!=620448401733239439360000. Go ahead and try out the following warm-up problem:
Evaluate ((22)!!)!.
Some Useful Theorems
For any non-negative integer n, we find that
n!!n!=(n−1)!! or n!=(n−1)!!×n!!.
We have the following 2 cases:
If n is odd, n!!n!=n×(n−2)×(n−4)×⋯5×3×1n×(n−1)×(n−2)×⋯×3×2×1. Since all the odd numbers n,n−2,n−4,…,5,3 get canceled, we are left with the equation n!!n!=(n−1)!!.
If n is even, n!!n!=n×(n−2)×(n−4)×⋯×4×2n×(n−1)×(n−2)×⋯×3×2×1. Since all the even numbers n,n−2,n−4,…,4,2 get canceled, we are left with the equation n!!n!=(n−1)!!.
Combining both cases, we find that for any non-negative integer n, n!!n!=(n−1)!!.□
Suppose that n!! is defined as follows:
n!!=⎩⎪⎪⎨⎪⎪⎧n×(n−2)×⋯×5×3×1n×(n−2)×⋯×6×4×21if n is odd;if n is even;if n=0,−1.
Then what is
6!!9!÷6!9!!?
For any non-negative integer n, we find that
(2n)!!(2n+1)!=(2n+1)!!.
Here, there is no need to consider two separate cases because it makes no difference whether n is odd or even.