Tag: zero factorial equals one

Why is 0! equals 1?

College Mathematics, Combinatorics, Questions and Quandaries

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 my Introduction to Combinations post, 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 A, B, 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.

 

 

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