# Why does 0! = 1? [duplicate]

Possible Duplicate:
Prove $0! = 1$ from first principles

Why does $0! = 1$?

All I know of factorial is that $x!$ is equal to the product of all the numbers that come before it. The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0! = 0$. I'm perplexed as to why I have to account for this condition in my factorial function (Trying to learn Haskell). Thanks.

• This can be answered by a simple google search. Mar 6, 2011 at 19:05
• Possible duplicate: math.stackexchange.com/questions/20969/…
– user856
Mar 6, 2011 at 19:07
• Dear Orbit, please refer to the question linked to above on this website. Mar 6, 2011 at 19:41

Mostly it is based on convention, when one wants to define the quantity $\binom{n}{0} = \frac{n!}{n! 0!}$ for example. An intuitive way to look at it is $n!$ counts the number of ways to arrange $n$ distinct objects in a line, and there is only one way to arrange nothing.

• Same question here as I have for Eric, why 1 way to arrange nothing, instead of 0 ways? Mar 6, 2011 at 19:12
• Can you give a argument as to why it -should- be 0 ways? Mar 8, 2011 at 18:46
• +1 This is a nice answer, it seems the factorial operation was created exactly to count the number of ways to arrange $n$ objects in a line. Nov 25, 2012 at 2:13

In a combinatorial sense, $$n!$$ refers to the number of ways of permuting $$n$$ objects. There is exactly one way to permute 0 objects, that is doing nothing, so $$0!=1$$.

There are plenty of resources that already answer this question. Also see:

http://en.wikipedia.org/wiki/Factorial#Definition

• Intending on marking as accepted, because I'm no mathematician and this response makes sense to a commoner. However, I'm still curious why there is 1 way to permute 0 things, instead of 0 ways. Mar 6, 2011 at 19:10
• If you want to do nothing, there is a way to do it, you just don't do it. But if you say there are 0 ways to do nothing, then you are implying that it is impossible to do nothing, which is of course not the case. This is how I look at it. Mar 6, 2011 at 19:37

It's because $n! = \prod_{0<k\le n} k.$ ($n!$ is the product of all numbers $1, 2,\dots n$) For $n = 0$ there isn't any number greater then 0 and lesser or equal to $n$, so the product is empty; the empty product is defined by convention as 1.

It has many reasons.

For example, we can have a power series: $e^x = \sum_{n} x^n/n!$ and we would like the first term to be $1$.

Also, how many permutations are there of $0$ numbers? Well, one.

We know that $\binom{n}{n}$ and $\binom{n}{0}= 1$. Thus $0! = 1$.

• The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately. We treat binomial coefficients like $\binom{5}{6}$ separately already; the theorem assumes $0 \leq k \leq n$. Mar 6, 2011 at 19:33