2021年11月29日 · For example: the factorial of zero i.e. an empty set ( doesn't occur) is 1. As the empty set can be arranged only in 1 way - i.e. by filling nothing. Now, let's take an example: 5 distinct seats. How many ways 5 distinct seats can be arranged? - 5! ways i.e. 120. So, basically, factorial gives us the arrangements.

$\begingroup$ @PeterTamaroff: The OP asked, "what does it actually mean to take the factorial of a complex number?" And this answer helpfully but tersely says that one way to extend factorial is using the gamma function, and tells what the gamma function is. $\endgroup$ –

2015年6月29日 · Since the factorial function is defined recursively, $(n+1)!=n! \cdot (n+1)$, your question boils down to whether or not the recurrence relation has a closed form solution, which it doesn't have. You want to be able to skip around calculating $1!$ through $9!$. However $10!$ is defined by $9!$, so there isn't a way of skipping the intermediate ...

It is a valid question to extend the factorial, a function with natural numbers as argument, to larger domains, like real or complex numbers. The gamma function also showed up several times as certain integrals, so mathematicians gave it a name and of …

2015年4月21日 · Is there a notation for addition form of factorial? $$5! = 5\times4\times3\times2\times1$$ That's pretty obvious. But I'm wondering what I'd need to use to describe. $$5+4+3+2+1$$ like the factorial $5!$ way. EDIT: I know about the formula. I want to know if there's a short notation.

The gamma function, shown with a Greek capital gamma $\Gamma$, is a function that extends the factorial function to all real numbers, except to the negative integers and zero, for which it is not defined. $\Gamma(x)$ is related to the factorial in that it is equal to $(x-1)!$.

2018年5月15日 · I am trying to find a simple factorial of all the preceding odd numbers. If $9$ were to be picked the equation would read $9\\times 7\\times 5\\times 3\\times 1$ (only odd numbers can be picked). Would...

$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. $. Otherwise this would be restricted to $0 <k

The Factorial of a Rational number is defined by the Gamma function. A link is in the comments. Since,

However, there is a continuous variant of the factorial function called the Gamma function, ...