Results 1 to 5 of 5

Thread: Value of a binomial

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    9

    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 $\displaystyle n<k$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Belgium
    Posts
    1,254
    Thanks
    24

    Re: Value of a binomial

    $\displaystyle \binom{n}{k}=0$ if $\displaystyle k<0$ or $\displaystyle n<k$, so indeed.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2011
    Posts
    9

    Re: Value of a binomial

    Quote Originally Posted by Siron View Post
    $\displaystyle \binom{n}{k}=0$ if $\displaystyle k<0$ or $\displaystyle n<k$, so indeed.
    Is there any instance not to be defined like the $\displaystyle 1/n$ when $\displaystyle n=0$?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2011
    Posts
    27

    Re: Value of a binomial

    Quote Originally Posted by grafan View Post
    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.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Renji Rodrigo's Avatar
    Joined
    Sep 2009
    From
    Rio de janeiro
    Posts
    38

    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 $\displaystyle {z \choose -n}=0\; \forall z, n>0 \in N$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Jul 15th 2010, 05:33 AM
  2. Replies: 1
    Last Post: Nov 12th 2009, 12:38 AM
  3. Replies: 1
    Last Post: Mar 11th 2009, 11:09 PM
  4. Neg Binomial and Binomial algebra relationship
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Feb 24th 2009, 03:12 PM
  5. Relation between Negative Binomial and Binomial Distros
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Nov 5th 2007, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum