Talk:N Choose k is not greater than n^k

You don't to impose limits on $k$ - this holds for all $k \in \Z$.

Also best to refer to positive integers for $n$ rather than natural numbers, i.e. $n \in \Z_{\ge 0}$, as integers are always better in an arithmetical context (the reasons are subtle and explained to a certain extent on some of the pages where these objects are defined from the axioms).

Having established that, the boundary conditions (i.e. where $n = 0, k \le 0, k > n$) all need to be reported on. --prime mover 18:27, 10 February 2012 (EST)


 * Seconded. --Lord_Farin 18:40, 10 February 2012 (EST)


 * GFP: From your previous (reverted) question: you don't need to consider negative factorials. Look at the definition for binomial coefficents for $k \le 0, k > n$. You specifically asked a question about it the other day. --prime mover 01:22, 12 February 2012 (EST)


 * That's why I undid the question, I figured that out :/ but thanks --GFauxPas 01:27, 12 February 2012 (EST)