# Why does 0! = 1? (Part 2)

This common question arises because $0!$ does not fit the usual definition for $n!$. Recall that, for positive integers, we have

$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$

$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

$3! = 3 \cdot 2 \cdot 1 = 6$

$2! = 2 \cdot 1 = 2$

$1! = 1$

In Part 1 of this series, I discussed descending down this lines with repeated division to define $0!$.

Here’s a second way of explaining why $0!=1$ that may or may not be as convincing as the first explanation. Let’s count the number of “words” that can made using each of the three letters A, B, and C exactly once. Ignoring that most of these don’t appear in the dictionary, there are six possible words:

ABC, ACB, BAC, BCA, CAB, CBA

With two letters, there are only two possible words: AB and BA.

With four letters, there are 24 possible words:

DABC, DACB, DBAC, DBCA, DCAB, DCBA.

Evidently, there are $4!$ different words using four letters, $3!$ different words using three letters, and $2!$ different words using two letters.

Why does this happen? Let’s examine the case of four letters. First, there are $4$ different possible choices for the first letter in the word. Next, the second letter can be anything but the first letter, so there are $3$ different possibilities for the second letter. Then there are $2$ remaining possibilities for the third letter, leaving $1$ possibility for the last.

In summary, there are $4 \cdot 3 \cdot 2 \cdot 1$, or $4!$, different possible words. The same logic applies for words formed from three letters or any other number of letters.

What if there are 0 letters? Then there is only 1 possibility: not making any words. So it’s reasonable to define $0!=1$.

It turns out that there’s a natural way to define $x!$ for all complex numbers $x$ that are not negative integers. For example, there’s a reasonable way to define $\left( \frac{1}{2} \right)!$, $\left(- \frac{7}{3} \right)!$ and even $(1+2i)!$. I’ll probably discuss this in a future post.

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