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I know the question sounds silly, however all I could find is the mathematical proof justifying the same but the convincing inference is still missing. My intent basically was to ask for the better comprehension of "there is only one way of arranging no objects." Please be polite.

ShubhGOYAL5
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    0! is just the empty product – Math137 Jan 06 '15 at 14:08
  • @math137 I think this is what makes the most sense. You can say things like "it's just how it's defined" or "the gamma function at $1$ is equal to $1$" but these aren't as intuitively satisfying as the empty product explanation. – Zubin Mukerjee Jan 06 '15 at 14:11
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    There is no "physical" meaning. But it makes sense logically for a number of useful and convenient reasons. Most of math is convenient notation to simplify theorems and their display. – Matthew Levy Jan 06 '15 at 14:13
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    Asking for physical substance of a mathematical fact seems like an invitation to give you a punch on the nose. But I won't, since you also asked to be polite (and because I cannot;-). – Marc van Leeuwen Jan 06 '15 at 14:22
  • 3! is the number of ways we can order three elements. This can be done in six ways, (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1), so 3! is 6. Similarly, we have (1,2) and (2,1) being the only ways to arrange 2 elements, so 2!=2. Similarly, (1) gives us 1!=1, and () gives us 0!=1. – Akiva Weinberger Jan 06 '15 at 15:21

4 Answers4

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Number of permutations of 0 elements...?

Martín-Blas Pérez Pinilla
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There are a couple of explanations.

  1. It's useful because it extends the equality $(n+1)! = n!(n+1)$ up to $n=0$
  2. The $\Gamma$ function, which is equal to $\Gamma(n) = (n-1)!$ for all $n\geq 2$, has a value of $1$ at $1$, so if we define $n! = \Gamma(n-1)$, then $0! = \Gamma(1) = 1$
  3. $n!$ is the number of ways you can order a set with $n$ elements, and there is exactly one way to order an empty set.
  4. As one of the commenters said, $0!$ is also the empty product and thus equal to $1$.
5xum
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Intuitively, the factorial of an integer $n$ is the "product of all the integers from $1$ until $n$" - so $1!=1$, $3!=1\cdot2\cdot3=6$, etc., as you know. If we use this definition for $0$, then we get what is called an "empty product" - the product of no numbers.

What should the empty product be? First, let's look at addition, since it's easier. When you take the sum of no numbers, the empty sum, it should be the additive identity: zero. This is clear when you consider sums such as

$$\sum_{\substack{(2k+1) \text{ is even} \\ k \,\in\, \mathbb{N}}} k$$

Since there are no such $k$, this sum is the empty sum, which is zero.

Similarly, the empty product should be the multiplicative identity, which is $1$. Thus $$0!=1$$

Zubin Mukerjee
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I think that $0!=1$ by definition: $(n+1)!=(n+1)\cdot n!$ needs a start value and this value is the only integer that makes $1!=1$, which make sense in combinatorics.

Lehs
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