Tag: conceptual barrier

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:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
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.

green line

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.

Why does 0! = 1? (Part 1)

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

Going from the bottom line to the top, we see that start at 1, and then multiply by 2, then multiply by 3, then multiply by 4, then multiply by 5. To get 6!, we multiply the top line by 6:

6! = 6 \cdot 5! = 6 \cdot 120 = 720.

Because they’re formed by successive multiplications, the factorials get large very, very quickly. I still remember, years ago, writing lesson plans while listening to the game show Wheel of Fortune. After the contestant solved the final puzzle, Pat Sajak happily announced, “You’ve just won $40,320 in cash and prizes.” My instantaneous reaction: “Ah… that’s 8!.” Then I planted a firm facepalm for having factorials as my first reaction. (Perhaps not surprisingly, I was still single when this happened.)

Back to 0!. We can also work downward as well as upward through successive division. In other words,

5! divided by 5 is equal to 4!.

4! divided by 4 is equal to 3!.

3! divided by 3 is equal to 2!.

2! divided by 2 is equal to 1!.

Clearly, there’s one more possible step: dividing by 1. And so we define 0! to be equal to 1! divided by 1, or

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

Notice that there’s a natural way to take another step because division by 0 is not permissible. So we can define 0!, but we can’t define (-1)!, (-2)!, \dots.

In Part 2, I’ll present a second way of approaching this question.