## Ask Professor Puzzler

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A student asked this question today: "Why is 0! (zero factorial) equal to one, instead of zero?"

Good question! Let's begin by making sure everyone knows what the "!" (factorial) notation means. n! means "the product of all the integers that lie between n and 1, inclusive."

Thus, 4! = 24, because 4! = 4(3)(2)(1) = 24.

6! = 6(5)(4)(3)(2)(1) = 720

The strange thing, though, is that 0! = 1, and that doesn't really seem to match our definition. After all, the integers between 0 and 1 inclusive are 0 and 1, and when you multiply them together, you get zero, not one!

Okay, so maybe our definition is flawed. We'll come back to that later.

The thing is, though, we don't want 0! to be equal to zero, because it's not *useful*. You see, we use factorials when we're calculating combinations of things, or when we're expanding a binomial to a power.

If you have something like (x + 1)^{5}, the n^{th} term in the expansion of that is (_{5}C_{n-1})x^{6-n}. That's the binomial theorem.

The problem is, that theorem doesn't work if we say that 0! = 0. Why? Because _{5}C_{0} = ^{5!}/_{(0!·5!)}, and if 0! is zero, then we have a division by zero problem! On the other hand, if we say that 0! = 1, then this works out perfectly to _{5}C_{0} = 1.

And really, if you think about it, that makes sense: If you have five objects, in how many different ways can you choose none of them? Uh, one!

We can see that 0! = 1 makes sense using patterns, too. Consider this:

^{7!}/_{6!} = 7^{6!}/_{5!} = 6^{5!}/_{4!} = 5^{4!}/_{3!} = 4^{3!}/_{2!} = 3^{2!}/_{1!} = 2

Now to continue this pattern, what do we need next?

^{1!}/_{0!} = 1

Solve this equation for 0!, and you get: 0! = 1.

Okay, so it makes sense with the combination notation to say 0! = 1, and we can even see from patterns that it must equal 1. So the real problem is our definition. So maybe we should reword our definition a little bit.

I like this way of saying it: *For all non-negative n, **n! is the product of 1 with all the positive integers less than or equal to n*.

Does that work? Sure! It keeps everything else the same, but since there no positive integers less than zero, we're left with 1.

That's one way of getting around it. Another way is to just say *For all non-negative n, **n! is the product of all the positive integers less than or equal to n*. This way of defining it forces us to use the "empty product" definition, which says that multiplying together zero factors gives a result of one.

Or you can define it recursively by saying : *0! = 1, and for all integer n > 0, n! = n(n - 1)!*.

Or, you can simply do this: *For all positive n, **n! is the product of 1 with all the positive integers less than or equal to n, and 0! = 1.*

This last one defines n! when n>0, and then gives a special definition for 0!.

No matter how you choose to define it, the real point is that mathematicians chose to define it to be one instead of zero simply because it was of practical use to do so!