Intuitive explanation of extended binomial coefficient

Question

We all are familiar with the following formula -

$$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!} \space\space \space ; \space \space n>r$$

This is the binomial formula where $n$ and $r$ are positive integers ($r$ can be $0$). However, recently I have been introduced to the Extended binomial theorem which increases the domain of $n$ to any real number.

Suppose $n \in \mathbb{R}$ then for every $x$ satisfying $|x| < 1$, we have $$ (1+x)^{n} = \sum_{r=0}^{\infty} \dbinom{n}r x^r \space \space \space \space ; \space \space r = 0,1,2...$$ and the extended binomial coefficient is given as $$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!}$$

This is the extended binomial theorem. I do understand the intuition behind the (so as to say) regular binomial coefficient. In simplest language, $\dbinom{n}r$ basically means number of ways to choose $n$ different objects taken $r$ at a time. But in the extended binomial theorem, $n$ can be any real number and $n<r$ is also possible. So, in such a case is there any such intuitive explanation as of the regular binomial coefficient ? Simply writing, let say $\dbinom{0.5}3$. What does this mean ?

Answer 1

For removing restriction on $n$ to be a positive integer, we can remove $n$ from factorial notation, since factorial are defined for integers only.

For that, we should simplify a little :

$$\binom{n}{r}=\frac{n!}{r!(n-r)!}=\frac{n\times(n-1)\times(n-2) \dots (n-(r-1))\times (n-r)!}{r!(n-r)!}$$ $$\boxed{\binom{n}{r}=\frac{n\times(n-1)\times(n-2) \dots (n-(r-2))\times(n-(r-1))}{r!}}$$

$$\text{This is the generalised defination of} ~\binom{n}{r} ~~\forall ~~n \in \mathbb R.$$

For example: $$\binom{0.5}{3}=\frac{0.5 \times (0.5-1)\times (0.5-2)}{3!}=\frac{1}{16}$$