## How to Use the Factorial Notation

We have had several discussions about the factorial notation, so I think this introduction is a bit late. However, it is important that you know these basic facts in order to perform calculations and understand better in later discussions.

The factorial of a non-negative integer $n$ is the product of all the positive numbers less than $n$. For example, the

$4! = 4 \times 3 \times 2 \times 1 = 24$.

and

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

In Introduction to Permutations, we have discussed that there are $n!$ ways to arrange $n$ distinct objects into a sequence. For instance, if we have 3 objects namely A, B, and C, then they can be arranged in $3! = 3 \times 2 \times 1 = 6$ ways. The arrangement are as follows:

$ABC, ACB, BAC, BCA, CAB, CBA$.

We have also learned some reasons why mathematicians chose the definition $0! = 1$. » Read more

## Why is 0! equals 1?

Many books will tell you that $0!$ equals $1$ is a definition. There are actually a few reasons why this is so – the two of which are shown below.

Explanation 1:

Based on mypost, we can conclude that $n$ taken $n$ at a time is equal to $1$. This means, that there is only one way that you can group $n$ objects from $n$ objects. For example, we can only form one group consisting of $4$ letters from AB, C and D using all the 4 letters.

From above, we know that the $\displaystyle{n \choose n} = 1$. But, $\displaystyle{n \choose n} = \displaystyle\frac{n!}{(n-n)!n!} = \displaystyle\frac{n!}{0!n!} = 1$. To satisfy the equation, $0!$ must be equal to $1$.

Explanation 2:

We can also use the fact that $n! = n(n-1)!$. Dividing both sides by $n$, we have $\displaystyle\frac{n!}{n} = (n-1)!$. If we let $n = 1$, we have $\displaystyle\frac{1!}{1} = (1 - 1)! \Rightarrow 1 = 0!$ which is what we want to show.

Explanation 3

We can also use the following pattern. We know that $n! = n(n-1)(n-2)(n-3) \cdots 3(2)(1)$ which means that

$n! = n(n-1)!$.

Dividing both sides of the equation by $n$, we have

$(n - 1)! = \frac{n!}{n}$

Using this fact, we can check the following pattern.

$4! = \displaystyle \frac{5!}{5} = \frac{(5)(4)(3)(2)(1)}{5} = 24$

$3! = \displaystyle \frac{4!}{4} = \frac{(4)(3)(2)(1)}{4} = 6$

$2! = \displaystyle \frac{3!}{3} = \frac{(3)(2)(1)}{(3)} = 2$

$1! = \displaystyle \frac{2!}{2} = \frac{(2)(1)}{(2)}$

Now, we go to $0!$

$0! = \displaystyle \frac{1!}{1} = 1$

As we can see from the 3 examples, $0! = 1$.