On Jun 4, 10:25�pm, Steven D'Aprano <ste...@remove.this.cybersource.com.au> wrote: > On Thu, 04 Jun 2009 09:47:05 -0700, Mensanator wrote: > > After all, everybody knows that for m items taken n at a time, the > > counts are > > > perm �w/repl = m**n > > comb �w/repl = (m+n-1)!/(n!(m-1)!) > > perm wo/repl = m!/(m-n)! > > comb wo/repl = m!/(n!(m-n)!) > > "Everybody" knows? Be careful with those sweeping generalizations. > Combinatorics is a fairly specialized area not just of programming but > mathematics as well.
I would expect that. That was supposed to be funny. :-) > > I've done a spot poll of four developers here (two juniors and two > seniors) and *not one* could give all four formulae correctly first go. > One guy didn't recognize the terms (although on being reminded, he got > two of the four formula). Another one, a self-professed maths-geek, > worked them out from first principles. I didn't know all of them either. I had to look them up in the source code of the combinatorics library I wrote for myself (I use that stuff a lot in my research). BTW, I really appreciate the links posted a while ago on how to search for text in .py files on Windows. Could not have found them otherwise. I must have a dozen copies of the module on each of 4 disks (two hard & 2 flash) but the comments containing those formulae was only in a single place. There was a subtle point. A function that would return such counts is fine - provided you don't have to actually count them! I'm sure Mr. Hettinger knows that. > > -- > Steven -- http://mail.python.org/mailman/listinfo/python-list
