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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.

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    $\begingroup$ There are some schools of thought that do define such a combination to equal 0. $\endgroup$ Commented Nov 9, 2018 at 16:59
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    $\begingroup$ Yes, Pascal's triangle is "by default" zero outside the range $[0,n]$ In row $n$. Likewise, $\binom n{-1}=0$. $\endgroup$
    – user65203
    Commented Nov 9, 2018 at 17:00

3 Answers 3

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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 $${n \choose n+1} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot1\cdot 0 }{(n+1)!} = 0.$$

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

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

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