# Thread: Value of a binomial

1. ## Value of a binomial

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

2. ## Re: Value of a binomial

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

3. ## Re: Value of a binomial

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

4. ## Re: Value of a binomial

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

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

So yes, $\displaystyle \binom n k = 0$ when $\displaystyle n<k$ 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 $\displaystyle n! = \prod_{i=1}^n i$, then we have $\displaystyle n! = 1 (\forall n\leq 0)$. But it's not really usual to use factorial on negative integers (beside 0).
(But you can expand the fatorial to $\displaystyle \mathbb R \backslash \mathbb Z^-$ with the $\displaystyle \Gamma$ Euler's map.)

5. ## Re: Value of a binomial

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

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

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

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

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

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

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