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I have been taking math and going through various forms of factorials for the purposes of probability, but I have not been able to find a conclusive reason as to how one accounts for 0! I am not sure if there is a way to solve it?

Xander Henderson
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AnanyaG
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    $0!=1$. Depending on how you have the factorial defined, this is either directly included in the definition (as is the case for the recursive definition of $0!=1$ and $n!=n\cdot (n-1)!$ for each $n\geq 1$) or noted because of the empty product equaling $1$, or it can be recognized combinatorially as there is precisely one bijection between the empty set and itself, namely the empty-function. – JMoravitz May 21 '18 at 18:40

2 Answers2

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There is only one permutation of zero objects.

thesmallprint
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Work recursively:

$$3!=\frac{4!}{4}=\frac{24}{4}=6$$ $$2!=\frac{3!}{3}=\frac 63 =2$$ $$1!=\frac{2!}{2}=\frac 22 =1$$ $$0!=\frac{1!}{1}=\frac 11=1$$

Rhys Hughes
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    And $(-1)!=0!/0=1/0=\infty$? – Dietrich Burde May 21 '18 at 18:44
  • Doesn't exist unless you perhaps use the $\Gamma$ function. – Rhys Hughes May 21 '18 at 18:45
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    Maybe $0!$ also doesn't exist, unless you use the Gamma function? – Dietrich Burde May 21 '18 at 18:46
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    I wouldn't comment if this answers the question on a higher math level. Dietrich may have a point there. But at the high school level and/or anywhere where the factorial is introduced for the first time and where $0!=1$ is needed to work with standard formulas (eg. Binomial distribution for stats people) that incorporate the factorial, this answer is great imo. – imranfat May 21 '18 at 18:50
  • @FlybyNight Well, did you upvote? I did becoause I am going to use this answer for my intro with factorials to my students. I have an old electronic calculator (with a cord) that gives $0!=0$. I show that to people. Then I explain that thanks to Richard Nixon, inflation started to take dominance and so then in math we declared $0!=1$. We both know this ain't true, but students believe me (at least they say...) – imranfat May 21 '18 at 18:57
  • @XanderHenderson if is a big word. Using $n! = n(n-1)!$ leaves one chasing ones tail: to understand $0!$, I need to understand $(-1)!$. If we use $n! = (n+1)!/n$ then we can make sense of this through permutations: if we know how many ways there are to arrange 5 people then we can find how many ways to arrange 4 people by deleting a person: $4! = 5!/5$. – Fly by Night May 21 '18 at 18:57
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    It points out a pattern that $(n-1)!=\frac{n!}{n}$ which can be seen to be true for all $n\geq 2$ whichever definition you were given. It does not however give any insight as to why this pattern should continue to work for other values of $n$ and in particular gives no insight as to why this pattern should work for $n=1$ or for $n=0$. If we were to give a finite answer to $(-1)!$ this pattern clearly breaks for $n=0$ so why should we expect it to work for $n=1$? Any answer to this question should in my opinion include definitions and from those definitions arrive at a conclusion. – JMoravitz May 21 '18 at 18:58
  • @FlybyNight You are responding to a comment that I deleted about 5 seconds after I posted it, but: THAT WAS MY POINT. You have to start with definitions. If you don't start with a definition of what you mean by the "factorial function", any answer can make sense. A search of this site for questions that ask "What is the next term in this sequence?" should make this clear. – Xander Henderson May 21 '18 at 19:03
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    Unfortunately, I don't think this is a very good answer. It replaces the recursive definition of the factorial by a completely circular way of thinking, without any motivation. It's also misleading for the reason that Dietrich Burde pointed out. And it's really not even necessary, as J Moravitz's comment and thesmallprint's answer show! –  May 21 '18 at 19:05
  • @T.Bongers It makes perfect sense. How many ways are there to arrange five people at a bus stop? $5\times 4 \times 3 \times 2 \times 1 = 5!$. How many ways are there of arranging four people? Well, delete one of the people from the last problem. There were five places that person could have taken, so there are $5!/5$ ways of arranging four people at a bus stop, i.e. $4! = 5!/5$. – Fly by Night May 22 '18 at 19:02