Negative Combination

Question

If given $0<r<n$, then $C_r^n=\frac{n!}{(n-r)!r!}$. But what if $r<0$ since $r!$ will fail?

Answer 1

If $r<0$, then the combinatorial interpretation of $C^n_r$ as the number of ways to choose $r$ items from a set of $n$ objects fails, so you must specify what you mean by this expression. Normally, we define it to be $0$. Think of this as meaning that the Pascal triangle outside the normal range consists of only entries, or to represent the fact that there's no way to choose "$-1$ elements" from any set, or just as a convention to make it convenient to allow us to write $$(1+x)^n=\sum_k \binom{n}{k}x^k$$ and have $k$ run through all integers with no problem---this is a very useful (convenient) property when working with generating functions, for instance.

Answer 2

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.