Talk:Number of Compositions

Can $$n$$ legitimately be zero in this case? If it is, we need to treat it as a separate case as $$\binom {n-1}{k-1}$$ is wrong when $$n = 0$$. IMO better to state that $$n > 0$$ (i.e. is strictly positive). --prime mover 06:41, 6 January 2011 (UTC)

Well since there aren't $$2^{0-1}=\frac12$$ total compositions of 0, I would say it doesn't hold for $$n=0$$... --Alec (talk) 21:02, 9 January 2011 (UTC)