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Can we say that $$\binom{n}{k}=\frac{n!}{k!(n-k)!}\boldsymbol 1_{\{0\leq k\leq n\}}\ \ ?$$

My definition of $\binom{n}{k}$ is the number of subset of $k$ element from a set of $n$ element. But my lecture only define it for $0\leq k\leq n$. But following this definition, since there are no subset of $k$ element for $k>n$ or $k<0$, to write

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}\boldsymbol 1_{\{0\leq k\leq n\}}$$ make sense or not really ? (because, I see nowhere a document with such a definition... but maybe it's not defined for $k\notin [0,n]\cap \mathbb N$ ?)

Todd
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  • Usually it is indeed convention that $\binom nk=0$ if $k\notin [0,n]\cap \mathbb N$ but it depends on the context – Maximilian Janisch Dec 30 '19 at 18:40
  • $k>n$ makes sense (binomial coefficient is $0$ by definition), but $k<0$ isn't a binomial coefficient anymore. I am aware of an extension , allowing $n$ to be arbitary real, but $k$ is still a non-negative integer. No idea whether there is a generalization with $k<0$ – Peter Dec 30 '19 at 18:40
  • It is correct, but it is less than the whole story. See my answer below. – Michael Hardy Dec 30 '19 at 19:47

3 Answers3

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It depends on context. In some cases it can be useful to generalize to $n$ and/or $k$ that are not positive integers, with $$ {n \choose k} = \frac{\Gamma(n+1)}{\Gamma(k+1) \Gamma(n-k+1)} $$ (and taking a limit if $n$ or $k$ is a negative integer).

Robert Israel
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I'm guessing that your bold-faced "1" indicates a characteristic function.

But the binomial coefficients can be defined with fractional and negative arguments. See this question and its accepted answer:

Binomial coefficients $1/2\choose k$

So I think the answer to your question is "no."

B. Goddard
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It is quite correct and sometimes useful to say that, for example, $\dbinom 4 6=0,$ i.e. the number of subsets of size $6$ of a set of size $4$ is $0.$ So what you write is correct.

For example, suppose a subcommittee of $6$ members is to be chosen from a committee of $4$ Republicans and $10$ Democrats. The number of such subcommittees is $\dbinom{14}6.$ The number of Democrats on the subcommittee belongs to the set $\{0,1,2,3,4,5,6\},$ and so does the number of Republicans, but that number cannot be $5$ or $6.$ Now suppose we want the probability that the Democrats are a majority of the subcommittee. The number of ways to get $4$ Democrats and $2$ Republicans is $\dbinom {10} 4\cdot \dbinom 4 2.$ The number of ways to get $5$ Democrats and $1$ Republicans is $\dbinom {10} 5\cdot \dbinom 4 1.$ And the number of ways to get $6$ Democrats and $0$ Republicans is $\dbinom {10} 6\cdot \dbinom 4 0.$ (And this last is an example of why the concept of the empty set makes sense: if the number of ways to get $0$ Republicans is not $1$ the we get a wrong answer here.) In the same way, the number of ways to get $5$ Republicans and $1$ Democrat is $\dbinom {10} 1\cdot\dbinom 4 5 = 0.$ And that shows up here: $$ \binom {14} 6 = \sum_{k=0}^6 \binom 4 k \cdot \binom {10} {6-k} \tag 1 $$ If we don't allow things like $\dbinom 4 5 =0,$ then identities like $(1)$ could not be presented without complicating things by talking about exceptions or about the narrow enough class of cases to which identities like this apply.

What you wrote is correct, but in some contexts, just as one writes $\dbinom 8 3 = \dfrac{8\times7\times6}{3\times2\times1},$ so one also writes $\dbinom {8.4} 3 = \dfrac{8.4\times7.4\times6.4}{3\times2\times1}.$ For example, this comes up in the binomial expansion: $$ (1+x)^{8.4} = 1 + 8.4x + \binom{8.4}2 x^2 + \binom {8.4}3 x^3 + \cdots \text{ for } |x|<1. $$ And negative integers in this context have a combinatorial interpretation: $$ \left| \binom{-8} 3 \right| $$ is the number of submultisets of size $3$ in a set of size $8.$

Michael Hardy
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