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Course: Precalculus (Eureka Math/EngageNY)>Unit 5

Lesson 2: Topic A: Lesson 2: Counting rules—The fundamental counting principle and permutations

Zero factorial or 0!

Learn all about factorials! Factorials are a quick way to represent multiplying a number by all the smaller positive integers down to one. This video also shows why mathematicians have defined zero factorial as one, instead of zero, to make the formula for permutations work in all cases.

Want to join the conversation?

• Is there any proof that 0 factorial equals 1?
• well i am in seventh grade so don't expect a great ans but i found that:
3!=4!/4
so 1!=2!/2 and
0!=1!/1
hope it helps
• Is it possible to have 2 factorial signs -- like 5!! ? Would that be 5! x 5! ?
• That would be something called a "multifactorial", which, for the case of two factorials is called a double factorial. For a number n, this is defined as:
n!! = n * (n-2) * (n-4) * ....

Basically, instead of subtracting 1 each time, we subtract 2 (or if we had n!!! we'd subtract 3, and so forth). If we have k factorial signs (and hence subtract k each time), the sequence of multiplication ends when we can't subtract k without getting a negative number.
• Can you give an example where 0! can actually be used? Like for example the chair arrangement? Thanks
• Yes. How many ways can 0 chairs be arranged in a row? Only 1 way – that is to do nothing at all!
• I think its also worth noting there is only one permutation of a set of zero items.

Here's an example:
• Guys I have an alternate way....
well we know,
n! = n(n-1)! as per defination of factorial.
so if we substitute n=1,we get,
1!=1(1-1)!
=> 1!=1x(0!)
hence, 0!=1 [as 1!=1]

Please tell me if this method could also be acceptable or not and do share your views...I was playing with formulae and accidentally came up with this...
have a good day
• Can we define n! = (n+1)!/(n+1), if we substitute 0 in the above formula we can get 0! = 1.
• Yes
You see (n+1)! = (n+1)x(n)x(n-1) ... , i.e. , (n+1)!= (n+1) x n!

Therefore (n+1)!/ (n+1) = (n+1) x n!/ (n+1) = (n+1) cancels out = n!
• if 0! is 1 and 1! is 1 by obvious logic 1 should be equal to 0. What are the terms and conditions for such a scenario when x! is equal to y!
• If x!=y!, it doesn't follow that x-y. I know that because otherwise 0 would equal 1 and it doesn't.
I would conclude that if x!=y! it follows that x=y unless x or y is 0 or 1.
I have never seen an equation where x! = y! so it doesn't come up much.
• Is this a correct reasoning:
Suppose you have n people and 0 chairs, there is only one combination possible that no one sits. so 0!=1
• Cant we do this concepts with rational number ? I ma thinking why cant N go down below zero ?
• The only formulas you have at your disposal at the moment is (n+1)! = (n+1) n! and 1! = 1. Using this with n=0, we would get 1! = (1)(0!) or 0! = 1!/1, so there's nothing too unnatural about declaring from that that 0! = 1 (and the more time you spend learning math, the more it will seem to be the correct choice intuitively). Now let's try the same trick to define (-1)!. Setting n = -1 in our formula above, we get 0! = (0) (-1)! or (-1)! = 0!/0. But now we're in undefined land, because you can't divide by zero, so the factorial function cannot be extended to negative integers.

Can you extend the factorial function to rational numbers (aside from the negative integers)? In theory, yes, but we don't have the tools in precalculus to talk about them. For instance, we could say that 1.5! = (1.5) 0.5!, but since we don't have any good ideas about what either of those factorials should be, we can't define any of them. When you get through calculus, you'll be able to understand a very awesome and weird function called the gamma function that actually accomplishes this task, but that's a conversation for another day. ^_^

https://en.wikipedia.org/wiki/Gamma_function