The value of factorial zero is equal to one. I have understood the mathematical interpretation of factorial zero. But how can I explain the meaning of factorial zero in common language
Asked
Active
Viewed 227 times
-2
-
4What exactly do you mean by common language? Maybe say it's like the fact that the empty product equals one? – Gregory Grant Jan 15 '16 at 13:08
-
It's also the only value such that the identity $n! = n (n - 1)!$, which is obvious for $n > 1$ from the usual definition, also holds for $n = 1$. I don't know whether this is "common language" in the sense you mean. – Travis Willse Jan 15 '16 at 13:11
-
1factorial counts the number of one to one, onto correspondences. There is only one such correspondence between two empty sets. – AlvinL Jan 15 '16 at 13:15
-
Yes sir. n!=n (n-1)(n-2)....3.2.1 where n is a natural number but zero is not a natural number. Then why 0!=1 – Achari S Ganesha Jan 15 '16 at 13:15
-
Whether or not $0 \in \mathbb{N}$ is a highly debatable issue. Some include it in $\mathbb{N}$, some do not include it in $\mathbb{N}$. – Clarinetist Jan 15 '16 at 13:20
-
1$0\in\mathbb{N}$? Blasphemy!!! Heretic!! – AlvinL Jan 15 '16 at 13:24
-
@AlvinLepik LOL – Clarinetist Jan 15 '16 at 13:30
-
@gebruiker: I'm not sure. The previous questions asks for a "proof from first principles"; this one seems to ask for an intuitive non-technical motivation. So even though the answers are more or less the same in both cases, the questions are still different. – hmakholm left over Monica Jan 15 '16 at 13:30
-
1Achari. My attention has been drawn to the fact that you hardly ever "accept" answers to your questions. Some users find it strange, offensive even, that you continue to ask for help, receive answers, but do not show your appreciation in any way. Read this meta thread for a relatively recent discussion on the matter. – Jyrki Lahtonen Jan 16 '16 at 08:02
3 Answers
0
In probability, we define $n!$ to be the number of possible orderings of $n$ objects, $n \geq 0$.
How many ways are there to order $0$ objects? $1$. Hence $0! = 1$.
See also this link.
Glorfindel
- 3,955
Clarinetist
- 19,519
-
IMO, one could argue that the number of ways to order $0$ objects is $0$. – barak manos Jan 15 '16 at 13:27
-
2It takes some subtlety to see that the empty relation is in fact a total order on the empty set -- it is how the definitions work out, but it is not immediately intuitively true unless one has some familiarity with vacuous truths. – hmakholm left over Monica Jan 15 '16 at 13:29
-
Please explain how the number of ways to arrange no objects is one. – Achari S Ganesha Jan 15 '16 at 16:19
-
0
I see 2 primary approaches to go about explaining this:
Using the definition. The definition of factorial n says that factorial of zero is 1 (see reference below). In this way, $0!=1$ is a definition and is not an argument to be proved or disproved.
The second approach is to use combinatorial arguments (e.g. @Clarinetist answer) or using other known relations such as the Gamma function.
Personally, I am comfortable with the first approach (the definition) one.
References:
NoChance
- 6,427
0
$3!$ is the number of ways to arrange three objects: \begin{align} &(1,2,3),(1,3,2),(2,1,3)\\ &(2,3,1),(3,1,2),(3,2,1) \end{align} So $3!=6$.
$2!$ is the number of ways to arrange two objects: $$(1,2),(2,1)$$ So $2!=2$.
$1!$ is the number of ways to arrange one object: $$(1)$$ So $1!=1$.
$0!$ is the number of ways to arrange zero objects: $$()$$ So $0!=1$.
Akiva Weinberger
- 23,062