Does Combinatorial where r is greater than n a valid operation? Inspired by this question on SE.SE I became curious about the result of a doing a combination function where the number being chosen is greater than the number of items in the set (Ex. 4C5).
Would the answer to such a function be an empty set or would it be defined as an invalid operation (kinda like trying to divide by 0)?
 A: There are generalizations; see this link. That said, in the case when $r>n>0$ are both integers, $\binom{n}{r}=0$ as expected by the combinatorial interpretation.
A: Combinatorially, for $n$ and $r$ non-negative integers, $\dbinom{n}{r}$ is the number of ways of choosing $r$ pairwise distinct objects from $n$ objects.  If $r\gt n,$ then there are no ways of choosing $r$ pairwise distinct objects from $n$ objects (because we don't have enough objects to find $r$ of them). So $\dbinom{n}{r}=0$ for $r\gt n.$
A: It depends on application.
Typically, one takes $\binom{n}{k} = 0$ whenever $k > n$, which makes good sense if one is summing binomial coefficients (the "exceptions" with $k > n$ won't affect the sum).
If one is multiplying binomial coefficients, I can imagine insisting that $\binom{n}{k} = 1$ whenever $k > n$ (in which case the "exceptions" with $k > n$ won't affect the product).
In either case, one could avoid considering the exceptional case by declaring the possibility meaningless or (best of all) carefully constraining the variables so that it does not arise.
A: For $r$ a nonnegative integer, the binomial coefficient $\binom zr$ is an $r^{\text{th}}$-degree polynomial in $z:$
$$\binom zr=\frac{z(z-1)(z-2)\cdots(z-r+1)}{r!}.$$
Its value is zero when $z=0,1,2,\dots,r-1.$ For example, $\binom45=0.$
