Would it be reasonable to define $\binom{n}{n+1} = 0$?

Question

Would it be reasonable to define $\binom{n}{n+1} = 0$?

My thinking is that it should be possible, since there are no ways to select $n+1$ items from a group of $n$ objects.

Answer 1

One of many equivalent definitions of Newton symbol is that $${n \choose k} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot(n-k+1)}{k!}.$$ Therefore $${n \choose n+1} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot1\cdot 0 }{(n+1)!} = 0.$$

Answer 2

The only sensible definition is zero. This can be formalised using the Gamma function; note $1/\Gamma(z)=0$ for $n=0,-1,-2,\ldots$.

Answer 3

Think this as number of ways to select $n+1$ balls form $n$ balls = $\boxed{0}$