How are lopsided binomials (eg $\binom{n}{n+1})?$ defined?

Question

For instance is $\binom{n}{n+1}=0$ always or something else?

Answer 1

If $n$ is a natural number, the combinatorial definition applies without problem. Count the number of selections of $n+1$ distinct elements among a set of $n$ elements. Since there are no such selections the binomial coefficient is zero.

A more general definition of binomial coefficients, in which only the lower index has to be a natural number, is $$ \binom nk=\frac{n(n-1)\ldots(n-k+1)}{k!}.\tag1 $$ Here it is not guaranteed that $\binom nk=0$ whenever $k>n$, although one sees again that this is true for $n,k\in\mathbf N$, since the numerator then contains a factor$~0$. However for $n<0$ or $n$ noninteger, one has $\binom nk\neq0$ for all $k\geq0$.

Usually one defines $\binom nk=0$ whenever $k$ is a negative integer, and it can be left undefined for noninteger $k$ (as there is rarely any use for such cases). Note that $(1)$ breaks the symmetry between $\binom nk$ and $\binom n{n-k}$, in that for $n\notin\mathbf N$ the two expressions are never simultaneously defined by $(1)$, and for those cases where the "negative integer $k$" clause defines one of them, the expressions will usually have different values. In particular this definition is different from any definition based on trying to push $\frac{n!}{k!(n-k)!}$beyond the combinatorial case, as that formula has symmetry baked into it. Usually such attempts give poor results, with subltle limit arguments needed even to get any value at all out of the formulas in the most useful cases (those with $n,k\in\mathbf Z$).

Answer 2

If you think of Pascal's triangle embedded into an infinite grid or matrix, then setting ${n \choose k}=0$ for $k>n$ or $k<0$ makes sense not only combinatorially (as others have argued) but also keeps Pascal's rule intact: $$ {n \choose k-1} + {n \choose k} = {n + 1 \choose k} $$