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

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.

• There are some schools of thought that do define such a combination to equal 0. Nov 9 '18 at 16:59
• Yes, Pascal's triangle is "by default" zero outside the range $[0,n]$ In row $n$. Likewise, $\binom n{-1}=0$.
– user65203
Nov 9 '18 at 17:00

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 $${n \choose n+1} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot1\cdot 0 }{(n+1)!} = 0.$$
The only sensible definition is zero. This can be formalised using the Gamma function; note $$1/\Gamma(z)=0$$ for $$n=0,-1,-2,\ldots$$.
Think this as number of ways to select $$n+1$$ balls form $$n$$ balls = $$\boxed{0}$$