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I added a definition of the binomial coefficient, in the comment section there is more info on how I ended with the doubt.

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edited Jun 21 at 16:22

How can $\binom{n}{k}=\frac{n\cdot (n-1)\cdot ... \cdot (n-k+1)}{k!}$, with $k<0$ or $k>n$ be equal to $0$?

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asked Jun 21 at 15:42

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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