I am thinking what values can n and r take in the following notation, $\binom{n}{r} = {n!\over r!(n-r)!}$ ?
I mean which one among the following is correct and why?
- $0≤r≤n$ where $r,n\in Z_0^+---------(1)$
- $0≤r≤n$ where $r,n\in Z^+---------(2)$
- $0<r≤n$ where $r,n\in Z_0^+---------(3)$
- $0<r≤n$ where $r,n\in Z^+---------(4)$
Furthermore, can you kindly say if $\binom{0}{0}$ exists. If it does exist, I can not find it holding a physical meaning even though we can obtain 1 by simplifying.
I would like to know which is correct for nPr from the above 4 connections I have written. 1, 2, 3 or 4 and why?
Thanks in advance.
From a combinatorial definition, one has that $\binom{n}{r}$ is the answer to the question of "How many $r$-element subsets are there of an $n$-element set?"
Under this definition and interpretation, one has that the question is invalid and makes no sense when $n$ is not a non-negative integer (There do not exist any sets with $-2$ number of elements of $\pi$ number of elements, etc...) and similarly makes no sense when $r$ is not a non-negative integer (similarly there exist no subsets of size $-2$ or size $\pi$ etc...)
As such, we either treat $\binom{n}{r}$ as being equal to zero or being undefined in the case that $n$ or $r$ is not a non-negative integer.
The counting question is perfectly valid from a syntactical standpoint when the values are both non-negative integers. Some of these will be easy to answer immediately as "there are none" for example the question of "How many $5$ element subsets are there of a $2$-element set?" There are no five-element subsets of a two-element set so the answer here is zero. The notation $\binom{2}{5}$ is perfectly valid in this case as notating the answer to this question and equals zero.
As to the question of $\binom{0}{0}$, this is also a perfectly valid question! The question being "How many zero-element subsets of a zero-element set exist?" The answer to this question... is one! There is in fact a zero-element subset of a zero-element set... Recall that $\emptyset\subseteq \emptyset$ is true. We have then $\binom{0}{0}=1$
We have for this interpretation the following results:
$\binom{n}{r}$ is
There also exists a purely algebraic interpretation. $\binom{n}{r}$ is the coefficient of the $X^r$ term in the expansion of $(1+X)^n$. We have the binomial theorem, that for non-negative integer $n$ we have $(1+X)^n = \sum\limits_{r=0}^n \binom{n}{r}X^r$. This is further generalized to complex $n$. The generalized binomial theorem is for $\alpha\in\Bbb C$ and $|X|<1$ we have
$$(1+X)^\alpha = \sum\limits_{r=0}^\infty \binom{\alpha}{r}X^r$$
where here $\binom{\alpha}{r} = \dfrac{\alpha\frac{r}{~}}{r!}$ where $\alpha\frac{r}{~}=\underbrace{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-r+1)}_{r~\text{terms in the product}}$ is the falling factorial. Note that here we do not require anything of $\alpha$ but we do still require $r$ to be a non-negative integer.
The values now of things such as $\binom{2}{5}$ and $\binom{\pi}{2}$ which made little sense before now do make sense. $\binom{2}{5}=\dfrac{2\cdot 1\cdot 0\cdot (-1)\cdot (-2)}{5!}=0$ and $\binom{\pi}{2}=\dfrac{\pi(\pi-1)}{2!}$
The generalized binomial coefficient gives the same results for all non-negative integers. You still have the same identities such as Paschal's identity. There are just now many more cases which give non-zero defined results.