May 31, 2026
Why does 0! equal 1 instead of 0?
I understand that factorial means multiplying down:
and
But then I see this:
and it feels wrong.
If there are zero numbers to multiply, why is the answer and not ? Is this just a definition mathematicians picked so formulas work, or is there an actual reason behind it?
I also saw someone say "there is one way to arrange nothing", but I do not fully get what that means.
3 answers157 views
5 comments
Mike JohnsonMay 31, 2026
I used to read it as "multiplying zero things should give zero." The arrangement explanation is the first one that made it feel less random.
Random RantJun 1, 2026
So the real trap is thinking factorial is only multiplication, when it is also counting arrangements?
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3 Answers
The cleanest reason is counting.
$n!$ counts the number of ways to arrange $n$ distinct objects.
For example, if you have three objects, there are
$$3! = 6$$
possible arrangements.
If you have one object, there is exactly one arrangement:
$$1! = 1.$$
Now ask the same question for zero objects:
How many ways are there to arrange nothing?
There is exactly one way: leave the arrangement empty.
That is why
$$0! = 1.$$
It is not saying "zero times something equals one." It is saying the empty arrangement is one valid arrangement.
This also keeps combinations correct:
$$\binom{n}{0} = \frac{n!}{0!(n-0)!} = 1.$$
There is exactly one way to choose zero objects from any set: choose nothing.
2 comments
Mike JohnsonJun 1, 2026
The line "not zero times something" is important. I think that is exactly where my intuition went wrong.
Sarah MitchellJun 1, 2026
So for arrangements, empty is still a valid outcome. That finally makes some sense.
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You can also get $0! = 1$ by walking backward through the factorial pattern.
For positive integers,
$$n! = n(n-1)!.$$
So:
$$3! = 3 \cdot 2!$$
$$2! = 2 \cdot 1!$$
$$1! = 1 \cdot 0!.$$
But $1! = 1$, so the last line becomes
$$1 = 1 \cdot 0!.$$
Therefore
$$0! = 1.$$
If we tried to set $0! = 0$, the pattern would immediately break at $1!$.
So it is a definition, but not an arbitrary one. It is the definition that preserves the counting meaning and the recursive factorial rule at the same time.
1 comment
Random RantJun 1, 2026
This backward pattern is probably the fastest exam explanation.
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Another useful phrase is "empty product."
The empty sum is $0$ because $0$ is the number that does not change addition:
$$a + 0 = a.$$
The empty product is $1$ because $1$ is the number that does not change multiplication:
$$a \times 1 = a.$$
So when a product has no factors, the natural neutral value is $1$, not $0$.
That idea is what makes many formulas behave cleanly. For example, in the binomial theorem:
$$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k.$$
The edge cases $k=0$ and $k=n$ only work smoothly when $0! = 1$.
This is also why students sometimes confuse $0!$ with $0^0$. They are related by "empty" reasoning in some contexts, but they are not the same expression.
