# Does $^nC_{n+1}$ exist? [duplicate]

is there any value assigned to $^nC_{n+1}$?

My teacher wrote it equal to $0$, but what will negative factorial mean?

$$^nC_{n+1} = \frac{n!}{(n+1)!\cdot(n-n-1)!} = \frac{n!}{(n+1)!\cdot(-1)!}$$

Thanks

## marked as duplicate by N. F. Taussig, Antonio Vargas, mfl, Ahaan S. Rungta, Andrew D. HwangMar 29 '15 at 21:05

• In several uses it is defined $\;\binom nk=0\;$ for $\;k>n\;$ . – Timbuc Mar 29 '15 at 11:54
The formula $$C(m,n)=\frac{m!}{n!(m-n)!}$$ is valid only if $m\ge n\ge 0$.
$C(m,m+1)$ is the number of subsets of $m+1$ elements in a set of $m$. Since there are no such subsets, $C(m,m+1)=0$.
A formula for ${}^{n}C_k$ valid for any complex number $n$ is $${}^n C_k = \frac{n(n-1)\dotsm(n-k+1)}{k!}.$$ If $n$ is an integer, and $k$ is an integer larger than $n$, the numerator contains the term $(n-(n+1-1)=0$, so the whole thing is $0$.
As I commented on in this answer, a much more useful general definition of binomials coefficients is $$\binom nk = \frac{n(n-1)\ldots(n-k+1)}{k!}, \qquad\text{for aribtrary n, and for k\in\Bbb N,}$$ possibly complemented by the convention the $\binom nk=0$ whenever $k$ is a negative integer (but such cases are much less often needed). This formula has no problem with having $k>n$, in which case it will give $0$ if (and only if) $n\in\Bbb N$, due to a factor $(n-n)=0$ in the numerator. This formula has the additional advantage of giving meaningful values when $n$ is negative or a fraction, or even a formal indeterminate. Many of the identities remain valid (but not the symmetry $\binom nk=\binom n{n-k}$), and one has Newtons generalised binomial formula. See also here and here.