Let $\ k,n\in\mathbb{N};\ k<n.\ $ Is there a nice, closed formula for the sum of products of $\ k\ $ numbers chosen from $\ \{1,\ldots,n\}\ ?$

Question

For example, when $\ n=5\ $ and $\ k=3\ $ we get

$$\ f(5,3) = (1)(2)(3) + (1)(2)(4) + (1)(2)(5) + (1)(3)(4) + (1)(3)(5) + (1)(4)(5) + (2)(3)(4) + (2)(3)(5) + (2)(4)(5) + (3)(4)(5) = 225.$$

But is there an nice, easy formula that spits out $\ 225\ $ if we give it $\ n=5\ $ and $\ k=3\ ?$ A formula that would be helpful for larger $\ n,k\ $ also.

This may or may not have to do with Newton's identities, which I saw mentioned in a vaguely related question.

Answer 1

These are the (unsigned) Stirling numbers of the first kind: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind . One definition of those is as the coefficients of the "rising factorial" polynomial $(x+1)(x+2) \cdots (x+n)$. There's not a nice closed formula but there are nice recurrences.

Answer 2

It would be the coefficient of $x^{n-k}$ of polynomial $p(x)=\prod_1^n(x+i)$, and we know the coefficient of polynomial is defined by $m$-th derivative of polynomial.

So there can be a form as...

$$ \frac1{(n-k)!}\frac {d^{n-k}}{dx^{n-k}}\prod_{i=1}^n(x+i)\Bigg|_{x=0} $$

I'm not sure it is closed form, but it is a form, at least.