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# Factorial also factors

Why is 0! = 1 ?

Note by Bodhisatwa Nandi
4 years, 3 months ago

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One way to explain this is similar to a way to explain why $$x^0=1$$ $$(x\neq 0)$$:

We know that $n!=n(n-1)!$ $1!=1(0)!$ $0!=1$

- 4 years, 3 months ago

the answer to this problem can be found by the gamma function which is defined as the integration of (e^-x).(x^m-1) where x varies from 0 to infinity; results in gamma m (m>0).. which is also equivalent to (m-1)! putting m=1 and solving the integration we get 0! = 1 ! abhhi bhool jaa engineering mein jayega tab indirectly iska proof mil jayega .. !! :)

- 4 years, 3 months ago

Thanks

- 3 years, 11 months ago

Ultimately it's all just a convention. When $$0! = 1$$, many things are simplified; for example, the binomial formula $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ holds true even for $$k=0,n$$, and it follows naturally from the identity $$n! = (n-1)! \cdot n$$ for $$n = 1$$. You can freely define it otherwise, but things become more complicated then (for example, the binomial formula works "for $$1 \le k \le n-1$$ only; if $$k = 0,n$$, the result is $$1$$").

The same thing applies for, for example, $$x^0 = 1$$ for nonzero $$x$$, or that $$1$$ is not a prime number, or that $$0$$ is even. They are just definitions and you're free to change them, but they generally make things more complicated to state if their definitions are changed.

- 4 years, 2 months ago