Why Does 0! = 1

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[Ninja]

This argument together with many other reasons is why 0!=1
1. 0! Represents an empty product which is conventionally taken as 1(similar to the 0th power for non-zero bases)
2. 0!=1 is the only definition that will stay consistent with our formulas such as nPr and nCr, as well as power series.
3. The gamma function(which is an extension of the factorial) also gives 1 when we input x=1 (gamma(x)=(x-1)!)

But I likr this argument the most because it’s the easiest to understand. “If you want to arrange nothing, there is only 1 way to do it, which is to put nothing.”

[Undertaker]

n! is also the number of permutations of n objects. Therefore, n! = P(n, n) = n!/(n - n)! = n!/0!. The only way these can be equal is if 0! = 1.

[Kiyoliki]

Here's why i thinks it's 1:
4! = 4 × 3 × 2 × 1
Divide it by 4, you get
3! = 3 × 2 × 1
Divide it by 3, you get
2! = 2 × 1
Divide by 2
1! = 1
Divide by 1
0! = 1

[chilldo]

The short actual (and not philosophical) explanation is like this: (n-1)! = n!/n. So 0! = (1-1)! = 1!/1 = 1/1 = 1

That's why there can't be negative factorials, not whole numbers at least:

(-1)! = (0-1)!= 0!/0 and oops, we're dividing by 0.

[jakobr_]

It may be unintuitive for some people that there is one way to arrange zero objects. If there’s nothing to arrange, surely there are no ways to arrange those things!

I think more reasoning is required.

Imagine we have two objects on the left, and three on the right. How many ways are there to arrange them, keeping both groups separate? 2!*3!.

Imagine we’re arranging 5 objects. 120 ways, right? Now consider the space after the last object. It’s empty, there are no objects there. We can pull these empty sets of objects out of anywhere as many times as we bother to! The number of ways to arrange 5 objects on the left and 0 objects on the right should be the same as a single set of 5 objects. Therefore, 0! is the multiplicative identity, 1.

[retro_alishka]

Bro you can do it with combination.
IT'S VERY EASY.
But thank you for video))

[tamirerez]

0! equal 1 because there is only 1 way to arrange 0 objects.

[seymenkapkiner]

I found this argument a bit confusing, wouldn’t that imply that 3!=7? Because you have 6 ways of arranging the numbers and additionally the option to just not arrange them at all ( if not arranging them at all is a actual case).

[jimmy]

If one of the ways to arrange is you don't, wouldn’t every factorial be nCk+1? (Where n is the number of objects, C is all the combinations you can arrange said objects, and k are the arranged objects from the source objects "n" you choose to arrange)?

Wouldn’t this change 2! from 2 to 3.
(1]One penny, one quarter, 2]one quarter, one penny, 3] and don't arrange them)?

[maricelty]

n! is also equal to n×(n-1!) so 0! must be 1 for 1! to be 1.

[angelcosta]

if you don't multiply a number, it's like multiplying by 1, so n^0=1
If you don't take a factorial, you didn't do anything, so you ''multiply by one yourself''
n!=1
Or that's what I thought

[xenphoton]

All things material/physical in this plane of existence, from cloud to Stone, should be considered more in nature of event rather than object🌝

[Blaqjaqshellaq]

1!=1, so 0! should equal 1!/1, which is 1.

[EmoBoyeww]

This is the worst explain i ever heard 😂

[thomasboi]

i have a proof
n! = n(n-1)!
1! = 1(1-1)!
divide both sides by 1
1 = 0!

[leesweets]

Its not a good argument.
How many ways are there to arrange -1 objects? One way, you dont. Ergo, all the negative integers factorial is defined to be 1.

[SleepTighty]

Because it doesn’t
0 times anything is 0