The definition of $\binom{n}{r}$ considering $r>n$ is allowed?

Question

After reading the post: Parameters of a Binomial Coefficient. I still have some confusion about the discrete definition(s) of $\binom{n}{r}$. I'm only interested in the case where $n, r\in\Bbb Z$, not the extended version of $\binom{n}{r}$ represented by Gamma function. (I would like to learn this approach when I'm more fluent at Calculus / continuous sort of things.)

I used to thought the definition is $\Large\binom{n}{r}:=\frac{n!}{r!(n-r)!}$, because it is more concise in a way that it has less combinatorial meaning components, i.e. only three: $n!, r!, (n-r)!$, than the falling factorial way $\Large\binom{n}{r} :=\frac{n(n-1)\cdots(n-r+1)}{r(r-1)\cdots1}$, which has $2r$ components, and it's hard to read a long formula with numbers scattered around. But the one I prefer can't define $\binom{3}{4}$ by the formula $\frac{3!}{4!(3-4)!}$, since you can't choose $4$ out of $3$ so it should be zero, but this would mean $\frac{1}{(-1)!}=0$, I consider this combinatorially meaningless. While the other one can define it properly: $\binom{3}{4}=\frac{3\cdot2\cdot1\cdot0}{4\cdot3\cdot2\cdot1}$.

So what's the formal definition of it that can deal with the case $r>n$ by formula, without define negative factorial to be zero? (Although I don't know whether a negative factorial would be useful in the future.)

Answer 1

The usual convention is that binomial coefficients which are "out of range" equal zero.

But there's another convention. If $k$ is a nonnegative integer, then $$\binom{n}k=\frac1{k!}n(n-1)\cdots(n-k+1)$$ for integers $n\ge k$. So we could define $\binom{x}k$ as the polynomial $$\binom{x}k=\frac1{k!}x(x-1)\cdots(x-k+1)$$ of degree $k$. Its zeros are $0,1,\ldots,k-1$.

With either convention, $\binom 34=0$.

Answer 2

For $n\ge0$, $\binom{n}{r}$ is the number of size-$r$ subsets of $\{1,\,\cdots,\,n\}$, regardless of $r$. If $0\le r\le n$, we can prove$$\binom{n}{r}=\binom{n}{n-r}=\frac{\prod_{j=n-r+1}^nj}{r!}.$$We can maintain this in an extension to $n<0$, viz.$$\binom{-m}{r}:=(-1)^m\binom{m+r-1}{r}=(-1)^m\binom{m+r-1}{m-1}$$for $m:=-n>0$. In particular, this is nonzero iff $r\ge0$. But to unite these cases, we can define $\binom{n}{r}$ more generally as the $x^r$ coefficient in $(1+x)^n$, as per the generalized binomial theorem:$$\binom{n}{r}:=[x^r](1+x)^n=[x^{-1}]\frac{(1+x)^n}{x^{r+1}}=\oint_{|z=1|}\frac{(1+z)^ndz}{2\pi iz^{r+1}}.$$

Answer 3

Personally I think it is best to define using the Gamma function, that is $${}_x \mathrm{C}_y=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$$ Since this takes care of the "negative factorial". For real $\alpha$ and an integer $k$, Wikipedia cites $${}_\alpha \mathrm{C}_k = \frac{\operatorname{Fact}(\alpha,k,\downarrow)}{k!}$$ With $\operatorname{Fact}(\alpha,k,\downarrow)$ being the falling factorial, $$\operatorname{Fact}(\alpha,k,\downarrow)=\prod_{j=0}^{k-1}(\alpha-j)$$ But it is easy to see that this is always $0$ if $\alpha \in \mathbb{N}$ and $k>\alpha$.

Answer 4

If you interpret $\binom n r$ combinatorially as the number of subsets of cardinality $r$ in a set of cardinality $n,$ then the expression is defined only when $n$ and $r$ are cardinalities, thus in $\{0,1,2,3,\ldots\}$ (where that set specified in $\{\text{braces}\}$ does or does not include cardinalities of infinite sets depending on what you want to do). In that case $\binom n r=0$ when $r>n$ because the number of subsets of cardinality $r$ in a set of cardinality $n$ is $0$ in that case.

There is also the use of these expressions in the expansion of powers of binomials, when the exponent is not an integer:

$$ (x+y)^n = \sum_{k=0}^\infty \binom n k x^k y^{n-k} \quad \text{if either } n\in\{0,1,2,\ldots\} \text{ or } \left| \frac x y \right| <1 $$ where $$ \binom n k = \frac{n(n-1)(n-2) \cdots (n-k+1)}{k!} \\ \text{ and $n$ need not be an integer and need not be positive.} $$ In this binomial theorem, the identity $\dbinom n k = \dfrac{n!}{k!(n-k)!}$ does not hold if factorials are defined only for nonnegative integers. If $\ge0,$ then one can still define factorials thus: $$ r! = \int_0^\infty x^r e^{-x}\,dx. $$ The idea of $\dfrac 1 {(-1)!}=0$ can make sense if one analytically continues this definition of factorial and then construes the expression to mean $\displaystyle \lim_{r\,\to\,-1} \frac 1 {r!}.$ You don't even need the concept of analytic continuation to do this if you use the identity $r!(r+1) = (r+1)!$ to define factorials of negative non-integers.

And for negative integer values of $n,$ one can say $n!=\infty$ provided one takes this to mean neither $+\infty$ nor $-\infty$ but rather the $\infty$ that is at both ends of the real line.

Answer 5

For integral (even complex) $n$ and integral $r$ the following definition holds:

\begin{align*} \binom{n}{r}= \begin{cases} \frac{n(n-1)\cdots (n-r+1)}{r!}=\frac{n^{\underline{r}}}{r!}&\quad r\geq 0\\ 0&\quad r<0 \end{cases} \end{align*}

See for instance formula (5.1) in chapter Binomial Coefficients of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.