I know that you can easily calculate $_5C_3$ to be $\frac{5!}{3!2!}=10$. However, what about $_5C_{-3}$, for example? It surely wouldn't work out using the combinations formula as it would involve a negative factorial. Do we just say something like $_5C_{-3}=0$? Google's built in calculator says so.
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1The binomial coefficient $\binom n k$ is defined as being $0$ when $k < 0$ or $k > n$. – Prime Mover Jun 05 '21 at 06:50
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@PrimeMover Though with the definition $\binom nk=\frac1{k!}\prod_{j=0}^{k-1}n-j$ the fact that $\binom nk=0$ for $n\in\Bbb N$ and $k>n$ is actually a theorem. – Jun 05 '21 at 06:55
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@PrimeMover I see. I was just split between whether it would be undefined or 0. – bravesheeptribe Jun 05 '21 at 06:56
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And $\binom nk\ne0$ for $n\notin\mathbb N$ (assuming $k\in\mathbb N$) regardless of whether $n$ is greater than $k$, less than $k$, negative, or imaginary. The $k^\text{th}$ degree polynomial $\binom zk$ has only $k$ zeroes, namely $z=0,1,2,\dots,k-1$. – bof Jun 05 '21 at 07:11
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1@Gae.S. Trouble with that is you have to be extra careful about how you define $k!$ for negative $k$. Not saying your statement is wrong as such, more that it raises extra complications that need more explanation to ensure it is rigorous. Also bear in mind that a vacuous product is taken to be $1$ not $0$. – Prime Mover Jun 05 '21 at 07:11
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How many ways can you pick a negative number of things from a set of objects?
It is impossible. There are none, so the value of the choose function is zero.
Nij
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I think this is a nice way to intuitively remember the result in case of negative k in the function $\binom {n}k$, but IMO its misleading and would lead to incorrect conclusion in case of negative n. For example ${-5 \choose 3}$ = -10, even though "it's impossible to pick 3 things from a set of -5 things" – Shreyans May 19 '24 at 19:40
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You can check this and this question for details about negative n in $\binom {n}k$. You can check this for details on negative k or case where n<k. – Shreyans May 19 '24 at 19:43