1. Value of a binomial

Can this notation be defined: $\binom{n}{k}$ with $n and if yes its value will be $0$?
Alternatively, can be a factorial of a negative number and $(n-k)! =0$ when $n?

2. Re: Value of a binomial

$\binom{n}{k}=0$ if $k<0$ or $n, so indeed.

3. Re: Value of a binomial

Originally Posted by Siron
$\binom{n}{k}=0$ if $k<0$ or $n, so indeed.
Is there any instance not to be defined like the $1/n$ when $n=0$?

4. Re: Value of a binomial

Originally Posted by grafan
Can this notation be defined: $\binom{n}{k}$ with $n and if yes its value will be $0$?
Alternatively, can be a factorial of a negative number and $(n-k)! =0$ when $n?
The definition of $\binom n k$ is not $\frac {n!} {(n-k)!k!}$.

The definition is the number of subsets with size k of a set of size n. It appears that if $n \geq k\geq 0$, then $\binom n k = \frac{n!}{(n-k)!k!}$.

So yes, $\binom n k = 0$ when $n because there is no subset of size k of a set of size n in this case.

For the second question, if factorial is defined by $n! = \prod_{i=1}^n i$, then we have $n! = 1 (\forall n\leq 0)$. But it's not really usual to use factorial on negative integers (beside 0).
(But you can expand the fatorial to $\mathbb R \backslash \mathbb Z^-$ with the $\Gamma$ Euler's map.)

5. Re: Value of a binomial

We can define ${z \choose n}=\frac{\prod\limits^{n-1}_{k=0}(z-k)}{n!}$

for $n \in N$ and $z \in R$ (or $z \in C$, complex numbers)

if $n>z \geq 0$ and $z \in N$ follows from that definition that ${z \choose n} =0$.

With that definition we can prove the identity
${z+1 \choose n}={z \choose n}+{z \choose n-1}$
we can use this identity to extend the definition of the binomial for n<0,
for example, put n=0

${z+1 \choose 0}={z \choose 0}+{z \choose -1}$

$1=1+{z \choose -1}$

so ${z \choose -1}=0$
by induction we can prove ${z \choose -n}=0\; \forall z, n>0 \in N$