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Math Help - Value of a binomial

  1. #1
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    Value of a binomial

    Can this notation be defined:  \binom{n}{k} with  n<k and if yes its value will be 0?
    Alternatively, can be a factorial of a negative number and  (n-k)! =0 when n<k?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Value of a binomial

    \binom{n}{k}=0 if k<0 or n<k, so indeed.
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  3. #3
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    Re: Value of a binomial

    Quote Originally Posted by Siron View Post
    \binom{n}{k}=0 if k<0 or n<k, so indeed.
    Is there any instance not to be defined like the 1/n when n=0?
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  4. #4
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    Re: Value of a binomial

    Quote Originally Posted by grafan View Post
    Can this notation be defined:  \binom{n}{k} with  n<k and if yes its value will be 0?
    Alternatively, can be a factorial of a negative number and  (n-k)! =0 when n<k?
    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<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 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.)
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  5. #5
    Junior Member Renji Rodrigo's Avatar
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    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
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