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I can't mathematically understand how $\binom{n}{k}$, with $k<0$ or $k>n$, can be equal to $0$.

The part that I don't understand is (when $k < 0$) $\frac{n!}{k!\cdot (n-k)!}$, but $k!$ is undefined.

Also, when $k > n$, $\frac{n!}{k!\cdot (n-k)!}$, but $(n-k)!$ is undefined.

Is there any mathematical proof or it's just logic based on what the binomial coefficient means?

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    $\begingroup$ "based on what the binomial coefficient means" Here we get into a chicken and egg sort of scenario. To be able to communicate with one another... we must first decide what the symbols $\binom{n}{k}$ should represent. How are we defining it? Algebraically? Combinatorially? Recursively? Some other way? $\endgroup$
    – JMoravitz
    Commented Jun 21 at 15:49
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    $\begingroup$ The most typical way to define it is combinatorially... at which $\binom{n}{k}$ is defined to be equal to the number of $k$-element subsets of an $n$-element set. There are no "negative-sized subsets" to any set... just as there are not subsets of size larger than the size of the parent set they came from... hence it equals zero. Any of that stuff about factorials comes after the fact and is only claimed to equal the binomial coefficient in those situations where it makes sense to talk about. $\endgroup$
    – JMoravitz
    Commented Jun 21 at 15:51
  • $\begingroup$ The question is raised because I was trying to do a proof by induction over $n$ of $\sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k} = 0$, with $n > 0$. I know that it can be easily proof by using the binomial theorem, but I wanted to practice the proof by induction. In the induction step I ended with $\sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k-1} + \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k} + ...$, the second term is part of my induction hyphotesis, and ended with the first term where the doubt raised when $k=0$. $\endgroup$
    – Benjamin Letelier Lazo
    Commented Jun 21 at 16:02
  • $\begingroup$ That doesn't address the important point JMoravitz is doing in the comments: How are defining $\binom{n}{k}$? $\endgroup$
    – jjagmath
    Commented Jun 21 at 16:04
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    $\begingroup$ If that is truly all you have to go on as a definition, then I can not prove this to you. The definition is bad and incomplete. I suggest taking the standard definition instead or extending your definition to include the extreme cases for $k$. $\endgroup$
    – JMoravitz
    Commented Jun 21 at 16:44

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$$ k<0 \lor k<n\\ \binom nk=\cfrac{n!}{k!(n-k)!} $$ If we just think about the mathematical definition and look at what happens when the gamma function takes a negative input. $$ n!=\Gamma(n+1)=\int_0^\infty t^n e^{-t}dt $$ Notice the gamma function by itself can never be equal $0$ except possibly at $n=-\infty$. But it can diverge and as we know $\lim_{x\to \infty}\frac 1 x=0$ So if we can have the denominator equal to infinity (positive or negative) this should be possible. By our theory if $k! = \pm\infty$: $$ \lim_{a\to k}\cfrac{n!}{a!(n-a)!}=0 $$ Take for example $n=1$ and $k=-1$ ($k<n \land k<0$): $$ \lim_{a\to -1}\frac{1}{a!(1-a!)}=\frac{1}{(-1)!\times 2}=0 $$

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