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Why is zero factorial equal to 1?

  1. Nov 13, 2005 #1
    Why is zero factorial equal to 1?
  2. jcsd
  3. Nov 13, 2005 #2
    As they say :rolleyes::
    *Edit: Sorry about that! Here's why:
    Three reasons, from my perspective:

    1) 0! is defined to be equal to one.

    2) (This will sound weird. Oh well, here goes) You know how
    [tex]\frac{{n!}}{n} = \left( {n - 1} \right)! [/tex]
    And so,
    [tex]n = 1 \Rightarrow \frac{{1!}}{1} = 0! = 1 [/tex]

    3) Consider combinatorials, for example. How many ways are there to choose [tex]k[/tex] objects out of a total quantity of [itex]n[/itex] objects?

    *We would use the expression:
    [tex] \frac{{n!}}{{k!\left( {n - k} \right)!}} [/tex]

    So, for example, there are 6 ways to choose 2 objects out of a set of 4.

    But what if [tex]k = 0[/tex], --->i.e., we choose nothing?

    [tex]\frac{{n!}}{{0!\left( {n - 0} \right)!}} = \frac{{n!}}{{0!n!}} = \frac{1}{{0!}} = 1\;({\text{or else we have a problem}}) [/tex]

    You see, there only one way to choose nothing. :smile:
    I usually just go with the first statement:

    -"0! is defined to be equal to one"- :biggrin:
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3


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    Since (n choose 0) gives the expression of (n choose n), because there's only one way to choose everything, there must be only one way to choose nothing.

    Of course consider, for n distinct objects, there are n! permutations of those objects. So it seems that for 0 objects there's 1 possible permutation.
  5. Nov 14, 2005 #4
  6. Nov 14, 2005 #5
    Is this the correct definition of factorial, or is it inconsistant with 0! ?

    [tex]x! = \prod_{n=1}^{x} n[/tex]
    Last edited: Nov 14, 2005
  7. Nov 14, 2005 #6


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    This is the* definition when x is a positive integer. With the usual convention that an empty product is 1, it is consistant with our usual definition for 0! when x=0. However, x=0 is usually left as a special case and explicitly defined as 0!=1.

    *there's more than one equivalent way to define factorial of course
  8. Nov 14, 2005 #7
    Thanks guys, I was going with the explanation that "0! is just equal to 1 because it is." But I thought about the definition of n! in the book of n(n-1)(n-2).....and I couldn't see exactly how it was so that 0! = 0. So I came here, thanks.
  9. Nov 14, 2005 #8
    It's an interesting question because you sort of have to look at it in the abstract. Also, there are a few other ways to rationalize 0! = 1.

    For instance, if 0! was anything other than 1, the cosine function wouldn't make any sense. Consider: f(x) = Cos(x) = x^0/0! + x^2/2! -x^4/4!... where x is a radian measure. So, if 0! was not equal to 1, then the first term in the series would not equal 1, and the Taylor series that derived it would be wrong, which would turn everything that we know about math upside down.
  10. Nov 14, 2005 #9


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    It would just change how we express the power series, it would not have any effect whatsoever on the mathematics. This is the same situation as adopting the convention x^0=1 for x=0, without this your power series as expressed doesn't make sense either. This is not a mathematically compelling reason for either convention/definition, rather a notational one.
  11. Nov 14, 2005 #10


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    Define A as a finite set, define B as the set of permuatations of A, then |A|! = |B|.

    This holds true even when |A|=0 as B in this case contains the empty function (and the empty function only), so you can see there is justification for definition 0! = 1.
  12. Nov 14, 2005 #11

  13. Nov 20, 2005 #12
    [tex](n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}[/tex]

    I know that the gamma function can be used to solve factorials for any number but are there any other special rules like this one?
  14. Nov 22, 2005 #13
    The special case 0! is defined to have value 0!==1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set emptyset).

    from the MathWorld
  15. Nov 23, 2005 #14

    matt grime

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    Step back, take a breath and try to think what the person who first came up with such Taylor/McLaurin series might have done if 0! weren't 1, because surely he or she wouldn't have used something that was undefined or unsuitable?
  16. Nov 25, 2005 #15


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    We should pin this thread, or one of the numerous others like unto it, to the top of the forum. Then again, why? After all: I dig the gamma function.
  17. Sep 25, 2007 #16

    25 < my age < 54. Start from the number of my age and toss a fair coin, plus one if it is a head and minus one otherwise, on average, you need to toss 210 times to hit the bound, i.e. 25 or 54. There are two integers satisfying the condition. Fortunately, the smaller one is my age.
  18. Sep 25, 2007 #17


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    And I'm sure the gamma function is fascinated by you.
  19. Jul 6, 2011 #18
    Re: Factorial!

    could you show me some basic properties of factorials? is factorial distributive over product?
  20. Jul 6, 2011 #19
    Re: Factorial!

    Can't you test that hypothesis yourself?

    What are 6!, 3!, and 2! equal to?

    What would 3!*2! be?
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