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On Sat, Aug 25, 2012 at 10:26 PM, And Rosta <and.rosta@hidden.email> wrote: > > > >>> N(n,m) = Sum from i=1 to min(n,m) of n!m! / i!(n-i)!(m-i)! > >> > >> You have n men and m women, each may enter into monogamous heterosexual > >> marriage or may remain single. How many different marriage patterns are > >> there? > > > > That's the above N(n,m) > > Righto. I don't understand well enough to actually do sums with it, but > that doesn't matter. Think of "i" as the number of marriages. The number of marriages can go from 1 up to the minimum between n and m, i.e. until you run out of either men or women. (I didn't include the case of no marriages. If you want to include that case too, you start the sum from 0 instead of 1. That just adds one case.) So the i-th term of the sum is the number of arrangements with i marriages. There are n!m! / i!(n-i)!(m-i)! possible arrangements with i marriages taken from n men and m women. ! is the factorial sign: n! = n*(n-1)*(n-2)*...*3*2*1 4! = 4*3*2*1 = 24 mu'o mi'e xorxes