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I was looking at the identity $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}, 1 \leq r \leq n$, so in the case $r = n$ we have $\binom{n}{n} = \binom{n-1}{n-1} + \binom{n-1}{n}$ that is $1 = 1 + \binom{n-1}{n}$ thus $\binom{n-1}{n} = 0$, am I right?

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Removed the "(which also says that $(-1)! = 0$)", which was a silly mistake :$ – Edgar Sánchez Jan 23 '13 at 5:48
up vote 6 down vote accepted

This is asking how many ways you can take $n$ items from $n-1$ items - there are none. So you are correct.

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OK, but then we really do need $(-1)!$ defined as 0, right? – Edgar Sánchez Jan 23 '13 at 5:18
Except for the part about $(-1)!=0$ --- I don't see where that comes from, and it's not the usual convention. – Gerry Myerson Jan 23 '13 at 5:19
It would come from $\binom{n-1}{n} = \frac{(n-1)!}{n! (n-1-n)!}$ – Edgar Sánchez Jan 23 '13 at 5:20
That would put zero in the denominator. You really need $(-1)!=\infty$. And, indeed, the gamma function has poles at the non-positive integers. – Thomas Andrews Jan 23 '13 at 5:29
@EdgarSánchez But then you divide by $(-1)!$ ;) You don't want to divide by 0... BTW $\Gamma(-1)=\infty$ which suggests that the natural choice for $(-1)!=\infty$... – N. S. Jan 23 '13 at 5:29

In my small treatize about "sums of like powers" (pg 14 ff) where I "re-invented" the integrals of the Bernoulli polynomials by a matrix-approach, I arrived at a formulation, where setting $ x=\binom{n}{n+1}$ to $ {1 \over (n+1)*(-1)!} \ne 0 $ was a meaningful thing, because it appeared in a formalism, where writing $ \frac 11b + \frac 12b + \frac13b + ... = (\frac 11 + \frac 12 + \frac 13 + ...)b = \frac {\zeta(1)}b = -1$ (where $b = \frac 1{(-1)!} $) allowed to arrive at the correct solution for the Bernoulli-/Zeta-polynomials. (We can argue, that $\lim _{\delta \to 0} { \zeta(1+\delta) \over \Gamma(\delta) } = 1$ and $\lim _{\delta \to 0} { \zeta(1-\delta) \over \Gamma(-\delta) } = -1$)

This is not a proof, that $ x=\binom{n}{n+1}$ need not always "be taken as zero", but it made me seriously thinking, how (and where else) such a setting actually makes sense.

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