Wouldn’t this also mean that 1!=0!, why is this true?
In: 383
Because “doing nothing” always count as one way to arrange stuff.
If I give you three objects [1,2,3], you can:
1. Do nothing -> [1,2,3]
2. Swap the first two -> [2,1,3]
3. Swap the last two -> [1,3,2]
4. Swap the first and last -> [3,2,1]
5. Take the last and add it before the first -> [3,1,2]
6. Take the first and add it after the last -> [2,3,1]
That’s 6 ways to arrange the three objects, so 3! = 6.
And as you see, the first way was “do nothing”. And you can always “do nothing”. Even if I give you zero object, you can “do nothing”. So when you count the arrangements, you will always get at least 1, corresponding to that “do nothing”. So 0! must be at least 1.
And obviously, if I give you zero objects, there is no other things to do than “do nothing”, so having 0! greater than 1 would be absurd. So 0! = 1.
In mathematics we can define things however we want to. That’s one of the beauties of it. We define things and explore the consequences, looking for patterns.
When we define things we are looking for things that are interesting, useful and consistent with all our other rules (ideally all three of them, but hopefully at least two.
We could define 0! to be something other than 1. That is a perfectly valid thing to do in maths. But it turns out not to be very useful, interesting or consistent.
Whereas defining 0! to be 1 lines up with some other neat patterns and rules, and works out quite well.
Others have explained the 0! = “number of ways to rearrange 0 people” part, so I’ll try to convince you that 0! shouldn’t be 0.
You can think of 2 lines of people, with 2 distinct people in first one and 3 in the second.
How many distinct way can you rearrange them, assuming that no one leaves their line? Well, as for the first line, there are 2! arrangements. As for the second one, the answer is 3!.
Then, since you can pair 1 way of arranging the first line with 1 way of arranging the second line to make 1 distinct arrangement of the 2 lines, the number of distinct arrangements is 2! × 3!.
Now, what if the first line is empty? Then the answer should be 0! × 3!. If 0! = 0, this number should be 0. But you should get 3! different arrangements when shuffling the second line, no? Then adding an empty first line shouldn’t suddenly make the number of arrangement 0?
Look just how convenient it is for the number of permutation of an empty set to be 1!
Everyone here’s talking about combinatorics, which is a useful way to visualize it intuitively, but the real, mathematically rigorous reason has to do with the mathematical definition of the factorial, which doesn’t have to do with combinations (though that’s one useful application).
Definitionally, a factorial is a product of successive terms. The key is what is the empty product? What is yule product of the integers in the empty set?
The answer is 1, the empty product is 1, because the multiplicative identity is 1, just like the empty sum is 0, because the additive identity is 0.
Because you think factorial as a multiplication while it is actually a division.
n! = (n+1)! / (n+1)
For n = 0;
0! = 1! / 1 = 1
I’ve been criticised for this view before, but I believe that 0!=1 because it’s mathematically convenient for it to do so, rather than because it really does for anything real.
I found this to be a simple and fun way to explain it: [Eddie Woo](https://youtu.be/X32dce7_D48?si=DY_KIDM5NRpX-h9j)
A lot of people trying to give answers that prove something…
The reason 0! = 1 is literally “because we said so”. It’s not like factorials exist as an entity outside of mathematics. Mathematics is not physics. Factorials are not something that mathematics tried to explain, it’s something that we created entirely, and in the process of doing so we define it in a way that’s useful for our purposes.
Proofs in mathematics only prove things about mathematics.
As seen in Terminator, “There’s no fate but what we make for ourselves”.
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