N Choose Negative Number is Zero
Jump to navigation
Jump to search
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.
$\blacksquare$