N Choose Negative Number is Zero

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Theorem

Let $n \in \Z$ be an integer.

Let $k \in \Z_{<0}$ be a (strictly) negative integer.

Then:

$\dbinom n k = 0$


Proof

From Pascal's Rule we have:

$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$

Thus it is sufficient to prove that:

$\forall n \in \Z: \dbinom n {-1} = 0$


So:

\(\ds \dbinom n {-1}\) \(=\) \(\ds \dbinom {n + 1} 0 - \dbinom n 0\) Pascal's Rule
\(\ds \) \(=\) \(\ds 1 - 1\) Binomial Coefficient with Zero
\(\ds \) \(=\) \(\ds 0\)

Hence the result.

$\blacksquare$