Can someone show me why the following works out? Thanks!
C(5,5) = (5!) ÷ (5!)(5-5)!
= (5!) ÷ (5!)(0)!
= 5x4x3x2x1 ÷ (5x4x3x2x1)... ?
The calculator says this equals 1.
This is one of the reasons:
In calculus, there is something known as the Gamma Function. It is defined as follows:
$\displaystyle \Gamma\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$
When evaluating $\displaystyle \Gamma\left(n\right)$, where $\displaystyle n\in\mathbb{N}$, it has this special property:
$\displaystyle \Gamma\left(n\right)=\left(n-1\right)!$
Now, let's see what happens when $\displaystyle n=1$
This means that $\displaystyle \Gamma\left(1\right)=\color{red}0!$
By definition, $\displaystyle \Gamma\left(1\right)=\int_0^{\infty}e^{-t}\,dt=\left.\left[-e^{-t}\right]\right|_0^{\infty}=\left[-e^{-\infty}-\left(-e^{0}\right)\right]=e^0=\color{red}1$
Thus, $\displaystyle \Gamma\left(1\right)=\color{red}\boxed{0!=1}$
You have already seen several excellent reasons why we define 0! = 1, but here is one more.
A fundamental recurrence relation for the factorial function is
$\displaystyle (n+1)! = (n+1) \; n!$
We would like this recurrence to hold even when $\displaystyle n = 0$. This requires
$\displaystyle 1! = 1 \cdot 0!$
$\displaystyle 1 = 1 \cdot 0!$
$\displaystyle 1 = 0!$