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What happens when in a combination, $n \choose k$, $n < k$? Does this result in a zero, or is this undefined?

RVC
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  • I'm trying to understand what $sum_{k=0}^N {k \choose m]$ results in. – RVC Jan 25 '18 at 09:52
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    If $n$ is a nonnegative integer and $k$ is an integer then: $\binom{n}{k}=0$ if $k\notin{0,1,\dots,n}$. Very convenient. Equations like $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$ stay valid. – drhab Jan 25 '18 at 10:03

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Not necessarily $0$. We have definition $$\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$$ So for example if we take $n = -3$ and $k = 2$, we have $$\binom{-3}{2} = \frac{(-3)\cdot(-4)}{2!} = 6$$ If you don't believe, see https://www.wolframalpha.com/input/?i=C(-3,2). Selection of objects is just an analogy for natural number values of $n$ and $k$, I suggest you to use the definition instead, for a more general result.

ArsenBerk
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    I think that the OP is referring to the natural numbers case, since he mentions "combination" is the question. Otherwise, we could also extend this to real values of $n,k$, as done in the binomial series, for example: $$(1+x)^a=\sum_{n=0}^\infty\binom{a}{n}x^n$$ – Vassilis Markos Jan 25 '18 at 10:36
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    Now that you mentioned it, I realized that you are right. But in any case, giving more general definitions is better in my opinion since it includes the natural number case but the other definition does not include this case. – ArsenBerk Jan 25 '18 at 10:46
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    From a scientific point of view, I totally agree. From a teaching point of view, I think it depends on the case - which is a piece of information that we cannot have at this time, unfortunately. But this seems more like a meta-discussion. Any way, I consider this a useful answer! :) – Vassilis Markos Jan 25 '18 at 10:48
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    @ΒασίληςΜάρκος: I know this is an old post, but you and ArsenBerk may both be interested in the fact that this natural number generalization elegantly fits the Newton series as the canonical basis elements for the forward-shift operator. And as a teacher I actually find it uncomfortable to attempt to teach the binomial series version as using the generalization of combinations. Even the formula is precisely the same, it seems to me that it is qualitatively different, much as the gamma function is different from the factorial function. But we all agree that it is a subjective meta-discussion! – user21820 Oct 15 '19 at 18:42
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    For reference, this is what I meant by "canonical basis elements for the forward-shift operator" but the comment has a limited size. – user21820 Oct 15 '19 at 18:43
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Hint:

Since $\binom{n}{k}$ is the the number of ways you can choose a subset of $k$ numbers out of $n$, in how many ways can you do such a choice when $n<k$?

P.S. Ask for further explanation, if needed.

Vassilis Markos
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Imagine this intuitively. The value: $$\binom{n} {k} $$ deals with the number of ways of choosing $k$ objects from $n$ objects.

Now, imagine $n<k$. How many ways of selecting $k$ objects are possible?

Yes, the answer is zero.

  • Thank you for you answer. Yes, your explanation of the intuition is sound, but I had to confirm. But how would you explain say, ArsenBerk's answer above? – RVC Jan 25 '18 at 10:03
  • Note that in Arsen Berk's answer where $n\lt k$ and $\binom{n}{k}$ is not $0$, $n$ is either not an integer, or is negative. Neither of these admit the combinatorial intuition given here. – robjohn Jan 25 '18 at 10:24
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For natural $k,$ the $k^{\text{th}}$ degree polynomial $\binom xk=\frac{x(x-1)\cdots(x-k+1)}{k!}$ is $0$ only for $x=0,1,\dots,k-1.$
You can easily verify the identity $$\binom{-x}k=(-1)^k\binom{x+k-1}k.$$ For example, $\binom{-2}n=(-1)^n\binom{n+1}n=(-1)^n(n+1).$
Thus the case $m=-2$ of the binomial theorem $$(1+x)^m=\sum_{n=0}^\infty\binom mnx^n$$ simplifies to $$(1+x)^{-2}=\sum_{n=0}^\infty\binom{-2}nx^n=\sum_{n=0}^\infty(-1)^n(n+1)x^n=1-2x+3x^2-4x^3+\cdots$$ When $x$ is a natural number, the polynomial $\binom xk$ is equal to the number of $k$-element subsets of an $x$-element set. This can be used to give bijective proofs (proofs by counting) of many combinatorial identities, such as Pascal's identity $$\binom xk=\binom{x-1}k+\binom{x-1}{k-1}\ \ (k=1,2,3,\dots).$$ Namely, you can prove it for natural values of $x$ by a counting argument, and then since both sides are polynomials you can conclude that the identity is valid for all values of $x.$

bof
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We have $n \choose r$. (The common notation is $n$ and $r$)

\begin{align} n \choose r & = \frac{n!}{(n-r)!r!}\\ \end{align}

$n \lt r \implies n-r < 0$. The factorial is not defined for negative numbers, so $n \choose r$ does not have an answer if $n < r$.

Landuros
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  • Why the downvote? Please explain. – Landuros Jan 25 '18 at 09:58
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    As long as $r \ge 0$, we can simplify the expression since $n!$ includes $(n-r)!$ as a multiplier. So there is no negative factorial if $r \ge 0$. Indeed, we have $\binom{-3}{2} = 6$ for example, by definition, although $-3 < 2$. – ArsenBerk Jan 25 '18 at 10:00
  • What is $(-3)!$? ($-3$ factorial)? Is that defined? It has no meaningful answer. See https://math.stackexchange.com/questions/927382/what-does-the-factorial-of-a-negative-number-signify – Landuros Jan 25 '18 at 10:02
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    Definition of $\binom{n}{k}$ is $\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$ so you don't have $(-3)!$. – ArsenBerk Jan 25 '18 at 10:03
  • @ArsenBerk Please, double check your information before speaking it. The formal definition of $n \choose r$ is$ \frac{n!}{(n-r)!r!}$, as listed here https://en.wikipedia.org/wiki/Binomial_coefficient – Landuros Jan 25 '18 at 10:12
  • @ArsenBerk Though I do see the correctness of your argument. Perhaps there are two ways to tackle this proof - yours and mine may both be equally correct. – Landuros Jan 25 '18 at 10:14
  • It is the definition when $n$ and $r$ are natural numbers as I am saying from the beginning and for natural numbers the definition I gave is equivalent to this one. – ArsenBerk Jan 25 '18 at 10:16
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    By the way your answer is not correct since you say it doesn't have an answer. For example $\binom{2}{3} = 0$ by convention/intuition as stated in other answers. – ArsenBerk Jan 25 '18 at 10:57